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Why does the Earth follow an elliptical trajectory rather than a parabolic one?
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I was taught that when the acceleration experienced by a body is constant, that body follows a parabolic curve. This seems logical because constant acceleration means velocity that is linear and position that is quadratic. This is what I learned from projectiles: Bodies are thrown with an initial velocity near the surface of the Earth, they experience constant velocity and the result is a parabolic curve.
Now that doesn't apply on the orbit of the Earth. The gravitational force can be thought of as constant since the distance fron Earth to Sun can be thought of as constant too, which by Newton’s second Law means the acceleration of Earth is also constant. Wouldn't that mean that the Earth should just follow a parabomic path?
Is there a mathematical proof (similar to the one I mentioned about projectiles) giving the elliptical orbit as a result?
My question is, in a word, why can't the Earth be treated as a ptojectile? And if it can them why doesn't it behave like one?
newtonian-mechanics newtonian-gravity orbital-motion projectile celestial-mechanics
New contributor
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add a comment |
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I was taught that when the acceleration experienced by a body is constant, that body follows a parabolic curve. This seems logical because constant acceleration means velocity that is linear and position that is quadratic. This is what I learned from projectiles: Bodies are thrown with an initial velocity near the surface of the Earth, they experience constant velocity and the result is a parabolic curve.
Now that doesn't apply on the orbit of the Earth. The gravitational force can be thought of as constant since the distance fron Earth to Sun can be thought of as constant too, which by Newton’s second Law means the acceleration of Earth is also constant. Wouldn't that mean that the Earth should just follow a parabomic path?
Is there a mathematical proof (similar to the one I mentioned about projectiles) giving the elliptical orbit as a result?
My question is, in a word, why can't the Earth be treated as a ptojectile? And if it can them why doesn't it behave like one?
newtonian-mechanics newtonian-gravity orbital-motion projectile celestial-mechanics
New contributor
$endgroup$
$begingroup$
Yes there is a mathematical proof. You just solve Newton equation of motion for a body in the gravitational field of the earth and you get a set of solutions which are elipses, circles and parabolas.
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– Žarko Tomičić
6 hours ago
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And also, acceleration is a vector and it is not a constant for the earth revolving arround the sun because that vector is always changing direction.
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– Žarko Tomičić
6 hours ago
3
$begingroup$
In the typical ballistic problem you describe Earth is basically considered as an infinite plane because of the relative dimensions of the bodies. If I recall correctly, elliptical trajectories arise in the two body problem when you consider the full expression for the gravitational force, that is, the inverse square law. By the way, there are other possible trajectories depending on the energy of the system (in fact, all conic sections). Check the Kepler problem for some info.
$endgroup$
– Lith
6 hours ago
add a comment |
$begingroup$
I was taught that when the acceleration experienced by a body is constant, that body follows a parabolic curve. This seems logical because constant acceleration means velocity that is linear and position that is quadratic. This is what I learned from projectiles: Bodies are thrown with an initial velocity near the surface of the Earth, they experience constant velocity and the result is a parabolic curve.
Now that doesn't apply on the orbit of the Earth. The gravitational force can be thought of as constant since the distance fron Earth to Sun can be thought of as constant too, which by Newton’s second Law means the acceleration of Earth is also constant. Wouldn't that mean that the Earth should just follow a parabomic path?
Is there a mathematical proof (similar to the one I mentioned about projectiles) giving the elliptical orbit as a result?
My question is, in a word, why can't the Earth be treated as a ptojectile? And if it can them why doesn't it behave like one?
newtonian-mechanics newtonian-gravity orbital-motion projectile celestial-mechanics
New contributor
$endgroup$
I was taught that when the acceleration experienced by a body is constant, that body follows a parabolic curve. This seems logical because constant acceleration means velocity that is linear and position that is quadratic. This is what I learned from projectiles: Bodies are thrown with an initial velocity near the surface of the Earth, they experience constant velocity and the result is a parabolic curve.
Now that doesn't apply on the orbit of the Earth. The gravitational force can be thought of as constant since the distance fron Earth to Sun can be thought of as constant too, which by Newton’s second Law means the acceleration of Earth is also constant. Wouldn't that mean that the Earth should just follow a parabomic path?
Is there a mathematical proof (similar to the one I mentioned about projectiles) giving the elliptical orbit as a result?
My question is, in a word, why can't the Earth be treated as a ptojectile? And if it can them why doesn't it behave like one?
newtonian-mechanics newtonian-gravity orbital-motion projectile celestial-mechanics
newtonian-mechanics newtonian-gravity orbital-motion projectile celestial-mechanics
New contributor
New contributor
edited 46 mins ago
Qmechanic♦
109k122041258
109k122041258
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asked 6 hours ago
Nader YouhannaNader Youhanna
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133
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Yes there is a mathematical proof. You just solve Newton equation of motion for a body in the gravitational field of the earth and you get a set of solutions which are elipses, circles and parabolas.
$endgroup$
– Žarko Tomičić
6 hours ago
$begingroup$
And also, acceleration is a vector and it is not a constant for the earth revolving arround the sun because that vector is always changing direction.
$endgroup$
– Žarko Tomičić
6 hours ago
3
$begingroup$
In the typical ballistic problem you describe Earth is basically considered as an infinite plane because of the relative dimensions of the bodies. If I recall correctly, elliptical trajectories arise in the two body problem when you consider the full expression for the gravitational force, that is, the inverse square law. By the way, there are other possible trajectories depending on the energy of the system (in fact, all conic sections). Check the Kepler problem for some info.
$endgroup$
– Lith
6 hours ago
add a comment |
$begingroup$
Yes there is a mathematical proof. You just solve Newton equation of motion for a body in the gravitational field of the earth and you get a set of solutions which are elipses, circles and parabolas.
$endgroup$
– Žarko Tomičić
6 hours ago
$begingroup$
And also, acceleration is a vector and it is not a constant for the earth revolving arround the sun because that vector is always changing direction.
$endgroup$
– Žarko Tomičić
6 hours ago
3
$begingroup$
In the typical ballistic problem you describe Earth is basically considered as an infinite plane because of the relative dimensions of the bodies. If I recall correctly, elliptical trajectories arise in the two body problem when you consider the full expression for the gravitational force, that is, the inverse square law. By the way, there are other possible trajectories depending on the energy of the system (in fact, all conic sections). Check the Kepler problem for some info.
$endgroup$
– Lith
6 hours ago
$begingroup$
Yes there is a mathematical proof. You just solve Newton equation of motion for a body in the gravitational field of the earth and you get a set of solutions which are elipses, circles and parabolas.
$endgroup$
– Žarko Tomičić
6 hours ago
$begingroup$
Yes there is a mathematical proof. You just solve Newton equation of motion for a body in the gravitational field of the earth and you get a set of solutions which are elipses, circles and parabolas.
$endgroup$
– Žarko Tomičić
6 hours ago
$begingroup$
And also, acceleration is a vector and it is not a constant for the earth revolving arround the sun because that vector is always changing direction.
$endgroup$
– Žarko Tomičić
6 hours ago
$begingroup$
And also, acceleration is a vector and it is not a constant for the earth revolving arround the sun because that vector is always changing direction.
$endgroup$
– Žarko Tomičić
6 hours ago
3
3
$begingroup$
In the typical ballistic problem you describe Earth is basically considered as an infinite plane because of the relative dimensions of the bodies. If I recall correctly, elliptical trajectories arise in the two body problem when you consider the full expression for the gravitational force, that is, the inverse square law. By the way, there are other possible trajectories depending on the energy of the system (in fact, all conic sections). Check the Kepler problem for some info.
$endgroup$
– Lith
6 hours ago
$begingroup$
In the typical ballistic problem you describe Earth is basically considered as an infinite plane because of the relative dimensions of the bodies. If I recall correctly, elliptical trajectories arise in the two body problem when you consider the full expression for the gravitational force, that is, the inverse square law. By the way, there are other possible trajectories depending on the energy of the system (in fact, all conic sections). Check the Kepler problem for some info.
$endgroup$
– Lith
6 hours ago
add a comment |
4 Answers
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Now that doesn't apply on the orbit of the Earth. The gravitational force can be thought of as constant since the distance fron Earth to Sun can be thought of as constant too
You are correct that the strength or magnitude of the sun's gravitational field is very similar over the length of the earth's orbit, but the direction is not. In a uniform gravitational field, the direction would be the same everywhere.
Over the path of the earth's orbit, the sun's gravitational field points in different directions. This significant difference from a uniform field means that the earth's orbit is quite far from a parabola.
$endgroup$
add a comment |
$begingroup$
The gravitational force can be thought of as constant since the distance fron Earth to Sun can be thought of as constant too, which by Newton’s second Law means the acceleration of Earth is also constant. Wouldn't that mean that the Earth should just follow a parabolic path ?
No, the gravitational force of the Sun on the Earth is not a constant, for two reasons:
- it is changing direction all the time, that is, it is always toward the Sun as the Earth (in the Sun's reference frame) revolves around it, and
- it is changing in magnitude as the Earth gets closer and farther away. This is because the kinetic energy of the Earth due to its orbital motion is not large enough to let it move in a circular orbit.
If the Earth moved parabolically around the Sun, it would not be in a closed orbit. It would pass by the Sun once and never return. That's because in order to have a parabolic orbit with Newtonian gravity,
$$|vec{F}|=frac{Gm_Em_S}{r^2},$$ the kinetic energy of Earth would be to large to stay in orbit.
Is there a mathematical proof (similar to the one I mentioned about projectiles) giving the elliptical orbit as a result ?
Yes, and it can be found in multiple places, usually in university sophomore level classical mechanics (and even engineering mechanic) books. See books by Symon, Marion, Beer & JOhnston, Barger & Olsson, Taylor, just to suggest a few. It is a standard derivation involving calculus, and it too long to detail here.
And actually, a projectile on Earth is following an elliptical path, too, around the center (roughly) of the Earth. We approximate the Newtonian gravity as a constant magnitude, constant direction force for small areas (like football fields), and get the parabolic shape, which is actually a good approximation of a short elliptical path.
$endgroup$
$begingroup$
"the kinetic energy of the Earth due to its orbital motion is not large enough to let it move in a circular orbit" I think you should also mention the gravitational potential energy there too.
$endgroup$
– PM 2Ring
6 hours ago
add a comment |
$begingroup$
Projectiles' paths are not actually parabolic near earth. A parabola does not imply constant total velocity. A parabola is constant velocity in one direction and acceleration in a perpendicular direction.
On the small scale of most ballistics problems, the earth is much larger than the trajectory of the projectile. On such a scale, the earth can be approximated as a plane. Whereas gravity pulls projectiles towards the center of the earth, and therefore is a direction that is changing for any moving object, this direction doesn't change from straight down, again on such small scales. The acceleration is straight down, there is no acceleration across, so the result, to close approximation, is parabolic. If you were to observe the path closely and it were able to "fall"through the earth, it would follow an elliptical path.
The actual equation is:
$$frac{1}{r}=c_0+c_1sin{theta}+c_2cos{theta}$$ where $r$ is the distance from the center of the earth.
The c's are constants depending on the mass of the earth, the angular momentum(a constant of the motion), and the mass of the projectile.
$r$ is the distance of the projectile from the center of the earth. Theta is the same theta from polar coordinates.
Let:
$L=mr^2dot{theta}$.
$L$ is the angular momentum of the projectile about the earth. $dot{theta}$ is the angular velocity along the trajectory.
The angular momentum is constant in time. Taking the derivatives of both sides gives you some information of the time evolution of the system.
If you set up the zero coordinate of your theta correctly, you can assume $c_1=0$.
Doing that, $c_2/c_0$ determines the eccentricity of your trajectory. This parameter tells you the shape of the projectiles path: Orbital Eccentricity
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If... it we’re to fall through the earth...: Oh I know, tempting. But not quite. Once you “fall through” you start to leave part of the mass of the earth behind; that changes the problem and the curve is no longer an ellipse. (Literally “leave behind”; under these idealized conditions, the spherical shell of the earth that is farther from the center of the earth than the bullet exerts no gravity at all and the force gets smaller the closer you get to the center)
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– Euro Micelli
57 mins ago
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Yes I'd forgotten about that. The trajectory above eatth is part of an ellipse which has the center of Earth at a focus. Below the surface, gravitational force is linear with radius. I'll need to factor that in.
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– R. Romero
50 mins ago
add a comment |
$begingroup$
I was taught that when the acceleration experienced by a body is
constant, that body follows a parabolic curve
That last part is wrong.
Under central force, the body will follow a conic section of some sort. A parabola is one type of conic section. An ellipse is another. A circle is yet another.
Whether you follow a circle, an ellipse or a parabola depends on the initial conditions - the amount of the force and the angular velocity.
But very generally, body's don't follow a parabola, but a conic.
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Conic sections are what you get from inverse-square central forces. That's actually true (to a good approximation) for the earth-sun system, but is not what the OP is describing when he says "when the acceleration experienced by a body is constant".
$endgroup$
– Henning Makholm
3 hours ago
$begingroup$
Note - a circle is a special case of an ellipse.
$endgroup$
– David White
1 hour ago
add a comment |
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4 Answers
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4 Answers
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Now that doesn't apply on the orbit of the Earth. The gravitational force can be thought of as constant since the distance fron Earth to Sun can be thought of as constant too
You are correct that the strength or magnitude of the sun's gravitational field is very similar over the length of the earth's orbit, but the direction is not. In a uniform gravitational field, the direction would be the same everywhere.
Over the path of the earth's orbit, the sun's gravitational field points in different directions. This significant difference from a uniform field means that the earth's orbit is quite far from a parabola.
$endgroup$
add a comment |
$begingroup$
Now that doesn't apply on the orbit of the Earth. The gravitational force can be thought of as constant since the distance fron Earth to Sun can be thought of as constant too
You are correct that the strength or magnitude of the sun's gravitational field is very similar over the length of the earth's orbit, but the direction is not. In a uniform gravitational field, the direction would be the same everywhere.
Over the path of the earth's orbit, the sun's gravitational field points in different directions. This significant difference from a uniform field means that the earth's orbit is quite far from a parabola.
$endgroup$
add a comment |
$begingroup$
Now that doesn't apply on the orbit of the Earth. The gravitational force can be thought of as constant since the distance fron Earth to Sun can be thought of as constant too
You are correct that the strength or magnitude of the sun's gravitational field is very similar over the length of the earth's orbit, but the direction is not. In a uniform gravitational field, the direction would be the same everywhere.
Over the path of the earth's orbit, the sun's gravitational field points in different directions. This significant difference from a uniform field means that the earth's orbit is quite far from a parabola.
$endgroup$
Now that doesn't apply on the orbit of the Earth. The gravitational force can be thought of as constant since the distance fron Earth to Sun can be thought of as constant too
You are correct that the strength or magnitude of the sun's gravitational field is very similar over the length of the earth's orbit, but the direction is not. In a uniform gravitational field, the direction would be the same everywhere.
Over the path of the earth's orbit, the sun's gravitational field points in different directions. This significant difference from a uniform field means that the earth's orbit is quite far from a parabola.
answered 6 hours ago
BowlOfRedBowlOfRed
18.4k22848
18.4k22848
add a comment |
add a comment |
$begingroup$
The gravitational force can be thought of as constant since the distance fron Earth to Sun can be thought of as constant too, which by Newton’s second Law means the acceleration of Earth is also constant. Wouldn't that mean that the Earth should just follow a parabolic path ?
No, the gravitational force of the Sun on the Earth is not a constant, for two reasons:
- it is changing direction all the time, that is, it is always toward the Sun as the Earth (in the Sun's reference frame) revolves around it, and
- it is changing in magnitude as the Earth gets closer and farther away. This is because the kinetic energy of the Earth due to its orbital motion is not large enough to let it move in a circular orbit.
If the Earth moved parabolically around the Sun, it would not be in a closed orbit. It would pass by the Sun once and never return. That's because in order to have a parabolic orbit with Newtonian gravity,
$$|vec{F}|=frac{Gm_Em_S}{r^2},$$ the kinetic energy of Earth would be to large to stay in orbit.
Is there a mathematical proof (similar to the one I mentioned about projectiles) giving the elliptical orbit as a result ?
Yes, and it can be found in multiple places, usually in university sophomore level classical mechanics (and even engineering mechanic) books. See books by Symon, Marion, Beer & JOhnston, Barger & Olsson, Taylor, just to suggest a few. It is a standard derivation involving calculus, and it too long to detail here.
And actually, a projectile on Earth is following an elliptical path, too, around the center (roughly) of the Earth. We approximate the Newtonian gravity as a constant magnitude, constant direction force for small areas (like football fields), and get the parabolic shape, which is actually a good approximation of a short elliptical path.
$endgroup$
$begingroup$
"the kinetic energy of the Earth due to its orbital motion is not large enough to let it move in a circular orbit" I think you should also mention the gravitational potential energy there too.
$endgroup$
– PM 2Ring
6 hours ago
add a comment |
$begingroup$
The gravitational force can be thought of as constant since the distance fron Earth to Sun can be thought of as constant too, which by Newton’s second Law means the acceleration of Earth is also constant. Wouldn't that mean that the Earth should just follow a parabolic path ?
No, the gravitational force of the Sun on the Earth is not a constant, for two reasons:
- it is changing direction all the time, that is, it is always toward the Sun as the Earth (in the Sun's reference frame) revolves around it, and
- it is changing in magnitude as the Earth gets closer and farther away. This is because the kinetic energy of the Earth due to its orbital motion is not large enough to let it move in a circular orbit.
If the Earth moved parabolically around the Sun, it would not be in a closed orbit. It would pass by the Sun once and never return. That's because in order to have a parabolic orbit with Newtonian gravity,
$$|vec{F}|=frac{Gm_Em_S}{r^2},$$ the kinetic energy of Earth would be to large to stay in orbit.
Is there a mathematical proof (similar to the one I mentioned about projectiles) giving the elliptical orbit as a result ?
Yes, and it can be found in multiple places, usually in university sophomore level classical mechanics (and even engineering mechanic) books. See books by Symon, Marion, Beer & JOhnston, Barger & Olsson, Taylor, just to suggest a few. It is a standard derivation involving calculus, and it too long to detail here.
And actually, a projectile on Earth is following an elliptical path, too, around the center (roughly) of the Earth. We approximate the Newtonian gravity as a constant magnitude, constant direction force for small areas (like football fields), and get the parabolic shape, which is actually a good approximation of a short elliptical path.
$endgroup$
$begingroup$
"the kinetic energy of the Earth due to its orbital motion is not large enough to let it move in a circular orbit" I think you should also mention the gravitational potential energy there too.
$endgroup$
– PM 2Ring
6 hours ago
add a comment |
$begingroup$
The gravitational force can be thought of as constant since the distance fron Earth to Sun can be thought of as constant too, which by Newton’s second Law means the acceleration of Earth is also constant. Wouldn't that mean that the Earth should just follow a parabolic path ?
No, the gravitational force of the Sun on the Earth is not a constant, for two reasons:
- it is changing direction all the time, that is, it is always toward the Sun as the Earth (in the Sun's reference frame) revolves around it, and
- it is changing in magnitude as the Earth gets closer and farther away. This is because the kinetic energy of the Earth due to its orbital motion is not large enough to let it move in a circular orbit.
If the Earth moved parabolically around the Sun, it would not be in a closed orbit. It would pass by the Sun once and never return. That's because in order to have a parabolic orbit with Newtonian gravity,
$$|vec{F}|=frac{Gm_Em_S}{r^2},$$ the kinetic energy of Earth would be to large to stay in orbit.
Is there a mathematical proof (similar to the one I mentioned about projectiles) giving the elliptical orbit as a result ?
Yes, and it can be found in multiple places, usually in university sophomore level classical mechanics (and even engineering mechanic) books. See books by Symon, Marion, Beer & JOhnston, Barger & Olsson, Taylor, just to suggest a few. It is a standard derivation involving calculus, and it too long to detail here.
And actually, a projectile on Earth is following an elliptical path, too, around the center (roughly) of the Earth. We approximate the Newtonian gravity as a constant magnitude, constant direction force for small areas (like football fields), and get the parabolic shape, which is actually a good approximation of a short elliptical path.
$endgroup$
The gravitational force can be thought of as constant since the distance fron Earth to Sun can be thought of as constant too, which by Newton’s second Law means the acceleration of Earth is also constant. Wouldn't that mean that the Earth should just follow a parabolic path ?
No, the gravitational force of the Sun on the Earth is not a constant, for two reasons:
- it is changing direction all the time, that is, it is always toward the Sun as the Earth (in the Sun's reference frame) revolves around it, and
- it is changing in magnitude as the Earth gets closer and farther away. This is because the kinetic energy of the Earth due to its orbital motion is not large enough to let it move in a circular orbit.
If the Earth moved parabolically around the Sun, it would not be in a closed orbit. It would pass by the Sun once and never return. That's because in order to have a parabolic orbit with Newtonian gravity,
$$|vec{F}|=frac{Gm_Em_S}{r^2},$$ the kinetic energy of Earth would be to large to stay in orbit.
Is there a mathematical proof (similar to the one I mentioned about projectiles) giving the elliptical orbit as a result ?
Yes, and it can be found in multiple places, usually in university sophomore level classical mechanics (and even engineering mechanic) books. See books by Symon, Marion, Beer & JOhnston, Barger & Olsson, Taylor, just to suggest a few. It is a standard derivation involving calculus, and it too long to detail here.
And actually, a projectile on Earth is following an elliptical path, too, around the center (roughly) of the Earth. We approximate the Newtonian gravity as a constant magnitude, constant direction force for small areas (like football fields), and get the parabolic shape, which is actually a good approximation of a short elliptical path.
answered 6 hours ago
Bill NBill N
10.2k12342
10.2k12342
$begingroup$
"the kinetic energy of the Earth due to its orbital motion is not large enough to let it move in a circular orbit" I think you should also mention the gravitational potential energy there too.
$endgroup$
– PM 2Ring
6 hours ago
add a comment |
$begingroup$
"the kinetic energy of the Earth due to its orbital motion is not large enough to let it move in a circular orbit" I think you should also mention the gravitational potential energy there too.
$endgroup$
– PM 2Ring
6 hours ago
$begingroup$
"the kinetic energy of the Earth due to its orbital motion is not large enough to let it move in a circular orbit" I think you should also mention the gravitational potential energy there too.
$endgroup$
– PM 2Ring
6 hours ago
$begingroup$
"the kinetic energy of the Earth due to its orbital motion is not large enough to let it move in a circular orbit" I think you should also mention the gravitational potential energy there too.
$endgroup$
– PM 2Ring
6 hours ago
add a comment |
$begingroup$
Projectiles' paths are not actually parabolic near earth. A parabola does not imply constant total velocity. A parabola is constant velocity in one direction and acceleration in a perpendicular direction.
On the small scale of most ballistics problems, the earth is much larger than the trajectory of the projectile. On such a scale, the earth can be approximated as a plane. Whereas gravity pulls projectiles towards the center of the earth, and therefore is a direction that is changing for any moving object, this direction doesn't change from straight down, again on such small scales. The acceleration is straight down, there is no acceleration across, so the result, to close approximation, is parabolic. If you were to observe the path closely and it were able to "fall"through the earth, it would follow an elliptical path.
The actual equation is:
$$frac{1}{r}=c_0+c_1sin{theta}+c_2cos{theta}$$ where $r$ is the distance from the center of the earth.
The c's are constants depending on the mass of the earth, the angular momentum(a constant of the motion), and the mass of the projectile.
$r$ is the distance of the projectile from the center of the earth. Theta is the same theta from polar coordinates.
Let:
$L=mr^2dot{theta}$.
$L$ is the angular momentum of the projectile about the earth. $dot{theta}$ is the angular velocity along the trajectory.
The angular momentum is constant in time. Taking the derivatives of both sides gives you some information of the time evolution of the system.
If you set up the zero coordinate of your theta correctly, you can assume $c_1=0$.
Doing that, $c_2/c_0$ determines the eccentricity of your trajectory. This parameter tells you the shape of the projectiles path: Orbital Eccentricity
$endgroup$
$begingroup$
If... it we’re to fall through the earth...: Oh I know, tempting. But not quite. Once you “fall through” you start to leave part of the mass of the earth behind; that changes the problem and the curve is no longer an ellipse. (Literally “leave behind”; under these idealized conditions, the spherical shell of the earth that is farther from the center of the earth than the bullet exerts no gravity at all and the force gets smaller the closer you get to the center)
$endgroup$
– Euro Micelli
57 mins ago
$begingroup$
Yes I'd forgotten about that. The trajectory above eatth is part of an ellipse which has the center of Earth at a focus. Below the surface, gravitational force is linear with radius. I'll need to factor that in.
$endgroup$
– R. Romero
50 mins ago
add a comment |
$begingroup$
Projectiles' paths are not actually parabolic near earth. A parabola does not imply constant total velocity. A parabola is constant velocity in one direction and acceleration in a perpendicular direction.
On the small scale of most ballistics problems, the earth is much larger than the trajectory of the projectile. On such a scale, the earth can be approximated as a plane. Whereas gravity pulls projectiles towards the center of the earth, and therefore is a direction that is changing for any moving object, this direction doesn't change from straight down, again on such small scales. The acceleration is straight down, there is no acceleration across, so the result, to close approximation, is parabolic. If you were to observe the path closely and it were able to "fall"through the earth, it would follow an elliptical path.
The actual equation is:
$$frac{1}{r}=c_0+c_1sin{theta}+c_2cos{theta}$$ where $r$ is the distance from the center of the earth.
The c's are constants depending on the mass of the earth, the angular momentum(a constant of the motion), and the mass of the projectile.
$r$ is the distance of the projectile from the center of the earth. Theta is the same theta from polar coordinates.
Let:
$L=mr^2dot{theta}$.
$L$ is the angular momentum of the projectile about the earth. $dot{theta}$ is the angular velocity along the trajectory.
The angular momentum is constant in time. Taking the derivatives of both sides gives you some information of the time evolution of the system.
If you set up the zero coordinate of your theta correctly, you can assume $c_1=0$.
Doing that, $c_2/c_0$ determines the eccentricity of your trajectory. This parameter tells you the shape of the projectiles path: Orbital Eccentricity
$endgroup$
$begingroup$
If... it we’re to fall through the earth...: Oh I know, tempting. But not quite. Once you “fall through” you start to leave part of the mass of the earth behind; that changes the problem and the curve is no longer an ellipse. (Literally “leave behind”; under these idealized conditions, the spherical shell of the earth that is farther from the center of the earth than the bullet exerts no gravity at all and the force gets smaller the closer you get to the center)
$endgroup$
– Euro Micelli
57 mins ago
$begingroup$
Yes I'd forgotten about that. The trajectory above eatth is part of an ellipse which has the center of Earth at a focus. Below the surface, gravitational force is linear with radius. I'll need to factor that in.
$endgroup$
– R. Romero
50 mins ago
add a comment |
$begingroup$
Projectiles' paths are not actually parabolic near earth. A parabola does not imply constant total velocity. A parabola is constant velocity in one direction and acceleration in a perpendicular direction.
On the small scale of most ballistics problems, the earth is much larger than the trajectory of the projectile. On such a scale, the earth can be approximated as a plane. Whereas gravity pulls projectiles towards the center of the earth, and therefore is a direction that is changing for any moving object, this direction doesn't change from straight down, again on such small scales. The acceleration is straight down, there is no acceleration across, so the result, to close approximation, is parabolic. If you were to observe the path closely and it were able to "fall"through the earth, it would follow an elliptical path.
The actual equation is:
$$frac{1}{r}=c_0+c_1sin{theta}+c_2cos{theta}$$ where $r$ is the distance from the center of the earth.
The c's are constants depending on the mass of the earth, the angular momentum(a constant of the motion), and the mass of the projectile.
$r$ is the distance of the projectile from the center of the earth. Theta is the same theta from polar coordinates.
Let:
$L=mr^2dot{theta}$.
$L$ is the angular momentum of the projectile about the earth. $dot{theta}$ is the angular velocity along the trajectory.
The angular momentum is constant in time. Taking the derivatives of both sides gives you some information of the time evolution of the system.
If you set up the zero coordinate of your theta correctly, you can assume $c_1=0$.
Doing that, $c_2/c_0$ determines the eccentricity of your trajectory. This parameter tells you the shape of the projectiles path: Orbital Eccentricity
$endgroup$
Projectiles' paths are not actually parabolic near earth. A parabola does not imply constant total velocity. A parabola is constant velocity in one direction and acceleration in a perpendicular direction.
On the small scale of most ballistics problems, the earth is much larger than the trajectory of the projectile. On such a scale, the earth can be approximated as a plane. Whereas gravity pulls projectiles towards the center of the earth, and therefore is a direction that is changing for any moving object, this direction doesn't change from straight down, again on such small scales. The acceleration is straight down, there is no acceleration across, so the result, to close approximation, is parabolic. If you were to observe the path closely and it were able to "fall"through the earth, it would follow an elliptical path.
The actual equation is:
$$frac{1}{r}=c_0+c_1sin{theta}+c_2cos{theta}$$ where $r$ is the distance from the center of the earth.
The c's are constants depending on the mass of the earth, the angular momentum(a constant of the motion), and the mass of the projectile.
$r$ is the distance of the projectile from the center of the earth. Theta is the same theta from polar coordinates.
Let:
$L=mr^2dot{theta}$.
$L$ is the angular momentum of the projectile about the earth. $dot{theta}$ is the angular velocity along the trajectory.
The angular momentum is constant in time. Taking the derivatives of both sides gives you some information of the time evolution of the system.
If you set up the zero coordinate of your theta correctly, you can assume $c_1=0$.
Doing that, $c_2/c_0$ determines the eccentricity of your trajectory. This parameter tells you the shape of the projectiles path: Orbital Eccentricity
answered 6 hours ago
R. RomeroR. Romero
69210
69210
$begingroup$
If... it we’re to fall through the earth...: Oh I know, tempting. But not quite. Once you “fall through” you start to leave part of the mass of the earth behind; that changes the problem and the curve is no longer an ellipse. (Literally “leave behind”; under these idealized conditions, the spherical shell of the earth that is farther from the center of the earth than the bullet exerts no gravity at all and the force gets smaller the closer you get to the center)
$endgroup$
– Euro Micelli
57 mins ago
$begingroup$
Yes I'd forgotten about that. The trajectory above eatth is part of an ellipse which has the center of Earth at a focus. Below the surface, gravitational force is linear with radius. I'll need to factor that in.
$endgroup$
– R. Romero
50 mins ago
add a comment |
$begingroup$
If... it we’re to fall through the earth...: Oh I know, tempting. But not quite. Once you “fall through” you start to leave part of the mass of the earth behind; that changes the problem and the curve is no longer an ellipse. (Literally “leave behind”; under these idealized conditions, the spherical shell of the earth that is farther from the center of the earth than the bullet exerts no gravity at all and the force gets smaller the closer you get to the center)
$endgroup$
– Euro Micelli
57 mins ago
$begingroup$
Yes I'd forgotten about that. The trajectory above eatth is part of an ellipse which has the center of Earth at a focus. Below the surface, gravitational force is linear with radius. I'll need to factor that in.
$endgroup$
– R. Romero
50 mins ago
$begingroup$
If... it we’re to fall through the earth...: Oh I know, tempting. But not quite. Once you “fall through” you start to leave part of the mass of the earth behind; that changes the problem and the curve is no longer an ellipse. (Literally “leave behind”; under these idealized conditions, the spherical shell of the earth that is farther from the center of the earth than the bullet exerts no gravity at all and the force gets smaller the closer you get to the center)
$endgroup$
– Euro Micelli
57 mins ago
$begingroup$
If... it we’re to fall through the earth...: Oh I know, tempting. But not quite. Once you “fall through” you start to leave part of the mass of the earth behind; that changes the problem and the curve is no longer an ellipse. (Literally “leave behind”; under these idealized conditions, the spherical shell of the earth that is farther from the center of the earth than the bullet exerts no gravity at all and the force gets smaller the closer you get to the center)
$endgroup$
– Euro Micelli
57 mins ago
$begingroup$
Yes I'd forgotten about that. The trajectory above eatth is part of an ellipse which has the center of Earth at a focus. Below the surface, gravitational force is linear with radius. I'll need to factor that in.
$endgroup$
– R. Romero
50 mins ago
$begingroup$
Yes I'd forgotten about that. The trajectory above eatth is part of an ellipse which has the center of Earth at a focus. Below the surface, gravitational force is linear with radius. I'll need to factor that in.
$endgroup$
– R. Romero
50 mins ago
add a comment |
$begingroup$
I was taught that when the acceleration experienced by a body is
constant, that body follows a parabolic curve
That last part is wrong.
Under central force, the body will follow a conic section of some sort. A parabola is one type of conic section. An ellipse is another. A circle is yet another.
Whether you follow a circle, an ellipse or a parabola depends on the initial conditions - the amount of the force and the angular velocity.
But very generally, body's don't follow a parabola, but a conic.
$endgroup$
$begingroup$
Conic sections are what you get from inverse-square central forces. That's actually true (to a good approximation) for the earth-sun system, but is not what the OP is describing when he says "when the acceleration experienced by a body is constant".
$endgroup$
– Henning Makholm
3 hours ago
$begingroup$
Note - a circle is a special case of an ellipse.
$endgroup$
– David White
1 hour ago
add a comment |
$begingroup$
I was taught that when the acceleration experienced by a body is
constant, that body follows a parabolic curve
That last part is wrong.
Under central force, the body will follow a conic section of some sort. A parabola is one type of conic section. An ellipse is another. A circle is yet another.
Whether you follow a circle, an ellipse or a parabola depends on the initial conditions - the amount of the force and the angular velocity.
But very generally, body's don't follow a parabola, but a conic.
$endgroup$
$begingroup$
Conic sections are what you get from inverse-square central forces. That's actually true (to a good approximation) for the earth-sun system, but is not what the OP is describing when he says "when the acceleration experienced by a body is constant".
$endgroup$
– Henning Makholm
3 hours ago
$begingroup$
Note - a circle is a special case of an ellipse.
$endgroup$
– David White
1 hour ago
add a comment |
$begingroup$
I was taught that when the acceleration experienced by a body is
constant, that body follows a parabolic curve
That last part is wrong.
Under central force, the body will follow a conic section of some sort. A parabola is one type of conic section. An ellipse is another. A circle is yet another.
Whether you follow a circle, an ellipse or a parabola depends on the initial conditions - the amount of the force and the angular velocity.
But very generally, body's don't follow a parabola, but a conic.
$endgroup$
I was taught that when the acceleration experienced by a body is
constant, that body follows a parabolic curve
That last part is wrong.
Under central force, the body will follow a conic section of some sort. A parabola is one type of conic section. An ellipse is another. A circle is yet another.
Whether you follow a circle, an ellipse or a parabola depends on the initial conditions - the amount of the force and the angular velocity.
But very generally, body's don't follow a parabola, but a conic.
answered 5 hours ago
Maury MarkowitzMaury Markowitz
4,7341628
4,7341628
$begingroup$
Conic sections are what you get from inverse-square central forces. That's actually true (to a good approximation) for the earth-sun system, but is not what the OP is describing when he says "when the acceleration experienced by a body is constant".
$endgroup$
– Henning Makholm
3 hours ago
$begingroup$
Note - a circle is a special case of an ellipse.
$endgroup$
– David White
1 hour ago
add a comment |
$begingroup$
Conic sections are what you get from inverse-square central forces. That's actually true (to a good approximation) for the earth-sun system, but is not what the OP is describing when he says "when the acceleration experienced by a body is constant".
$endgroup$
– Henning Makholm
3 hours ago
$begingroup$
Note - a circle is a special case of an ellipse.
$endgroup$
– David White
1 hour ago
$begingroup$
Conic sections are what you get from inverse-square central forces. That's actually true (to a good approximation) for the earth-sun system, but is not what the OP is describing when he says "when the acceleration experienced by a body is constant".
$endgroup$
– Henning Makholm
3 hours ago
$begingroup$
Conic sections are what you get from inverse-square central forces. That's actually true (to a good approximation) for the earth-sun system, but is not what the OP is describing when he says "when the acceleration experienced by a body is constant".
$endgroup$
– Henning Makholm
3 hours ago
$begingroup$
Note - a circle is a special case of an ellipse.
$endgroup$
– David White
1 hour ago
$begingroup$
Note - a circle is a special case of an ellipse.
$endgroup$
– David White
1 hour ago
add a comment |
Nader Youhanna is a new contributor. Be nice, and check out our Code of Conduct.
Nader Youhanna is a new contributor. Be nice, and check out our Code of Conduct.
Nader Youhanna is a new contributor. Be nice, and check out our Code of Conduct.
Nader Youhanna is a new contributor. Be nice, and check out our Code of Conduct.
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Yes there is a mathematical proof. You just solve Newton equation of motion for a body in the gravitational field of the earth and you get a set of solutions which are elipses, circles and parabolas.
$endgroup$
– Žarko Tomičić
6 hours ago
$begingroup$
And also, acceleration is a vector and it is not a constant for the earth revolving arround the sun because that vector is always changing direction.
$endgroup$
– Žarko Tomičić
6 hours ago
3
$begingroup$
In the typical ballistic problem you describe Earth is basically considered as an infinite plane because of the relative dimensions of the bodies. If I recall correctly, elliptical trajectories arise in the two body problem when you consider the full expression for the gravitational force, that is, the inverse square law. By the way, there are other possible trajectories depending on the energy of the system (in fact, all conic sections). Check the Kepler problem for some info.
$endgroup$
– Lith
6 hours ago