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What happens when the drag force exceeds the weight of an object falling into earth?

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2

$begingroup$

Let's say a meteor is coming towards earth. It's not accelerating, but it does have an initial velocity. This meteor is shaped so it has an insane amount of drag, enough to even exceed its weight (not mass) as it gets closer. What happens? Why?

At first I thought it would just stop, but that doesn't really make sense. The forces cancel but that doesn't mean the body slows down. Would it keep going in another direction?

Thanks in advance!

gravity acceleration drag

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asked 1 hour ago

Laura IglesiasLaura Iglesias

111

New contributor

Laura Iglesias is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

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add a comment | 

2

$begingroup$

Let's say a meteor is coming towards earth. It's not accelerating, but it does have an initial velocity. This meteor is shaped so it has an insane amount of drag, enough to even exceed its weight (not mass) as it gets closer. What happens? Why?

At first I thought it would just stop, but that doesn't really make sense. The forces cancel but that doesn't mean the body slows down. Would it keep going in another direction?

Thanks in advance!

gravity acceleration drag

share|cite|improve this question

asked 1 hour ago

Laura IglesiasLaura Iglesias

111

New contributor

Laura Iglesias is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

add a comment | 

$begingroup$

Let's say a meteor is coming towards earth. It's not accelerating, but it does have an initial velocity. This meteor is shaped so it has an insane amount of drag, enough to even exceed its weight (not mass) as it gets closer. What happens? Why?

At first I thought it would just stop, but that doesn't really make sense. The forces cancel but that doesn't mean the body slows down. Would it keep going in another direction?

Thanks in advance!

gravity acceleration drag

share|cite|improve this question

asked 1 hour ago

Laura IglesiasLaura Iglesias

111

New contributor

Laura Iglesias is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

share|cite|improve this question

asked 1 hour ago

Laura IglesiasLaura Iglesias

111

New contributor

Laura Iglesias is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|cite|improve this question

asked 1 hour ago

Laura IglesiasLaura Iglesias

111

New contributor

Laura Iglesias is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked 1 hour ago

Laura IglesiasLaura Iglesias

111

New contributor

Laura Iglesias is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked 1 hour ago

Laura IglesiasLaura Iglesias

111

2 Answers
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2

$begingroup$

Drag is proportional to the object's velocity, so as the object's speed decreases, the drag decreases as well. Therefore, the body won't start "falling upwards" because for that to happen, the velocity must reach zero, and at that point the drag also drops to zero.

In practice, the falling object slows down until the drag equals the weight, at which point it keeps moving at constant speed (so-called "terminal velocity"). In the hypothetical scenario where you have a constant force that's greater than the weight pulling on the object, then the object will indeed start falling upwards.

share|cite|improve this answer

edited 39 mins ago

answered 1 hour ago

AllureAllure

2,351926

$endgroup$

add a comment | 

1

$begingroup$

The acceleration (or deceleration) of an object is $a$: $$a = f/m.$$

Acceleration (or deceleration) is the rate of change of velocity. So, to find the rate of change of velocity of an object you divide all the forces acting on it, by the object's mass.

The forces acting on your hypothetical object are the object's weight and the atmospheric drag. Someone being very picky might also say that there is a relatively very small atmospheric buoyancy at work too, but it can be ignored in most cases.

Let's say the object is falling straight down. Atmospheric drag is velocity-dependent, with low drag at low speeds and much higher drag at higher speeds. Drag will cause the object to decelerate, until drag equals the object's weight. At that point the object is said to be moving at terminal velocity, and it just keeps falling at terminal velocity.

The whole scenario gets more complicated when altitude-dependent atmospheric density is taken into account, but what's described above captures the essence of an answer to your question.

share|cite|improve this answer

answered 50 mins ago

S. McGrewS. McGrew

9,81421239

$endgroup$

add a comment | 

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2

$begingroup$

Drag is proportional to the object's velocity, so as the object's speed decreases, the drag decreases as well. Therefore, the body won't start "falling upwards" because for that to happen, the velocity must reach zero, and at that point the drag also drops to zero.

In practice, the falling object slows down until the drag equals the weight, at which point it keeps moving at constant speed (so-called "terminal velocity"). In the hypothetical scenario where you have a constant force that's greater than the weight pulling on the object, then the object will indeed start falling upwards.

share|cite|improve this answer

edited 39 mins ago

answered 1 hour ago

AllureAllure

2,351926

$endgroup$

add a comment | 

2

$begingroup$

Drag is proportional to the object's velocity, so as the object's speed decreases, the drag decreases as well. Therefore, the body won't start "falling upwards" because for that to happen, the velocity must reach zero, and at that point the drag also drops to zero.

In practice, the falling object slows down until the drag equals the weight, at which point it keeps moving at constant speed (so-called "terminal velocity"). In the hypothetical scenario where you have a constant force that's greater than the weight pulling on the object, then the object will indeed start falling upwards.

share|cite|improve this answer

edited 39 mins ago

answered 1 hour ago

AllureAllure

2,351926

$endgroup$

add a comment | 

$begingroup$

Drag is proportional to the object's velocity, so as the object's speed decreases, the drag decreases as well. Therefore, the body won't start "falling upwards" because for that to happen, the velocity must reach zero, and at that point the drag also drops to zero.

In practice, the falling object slows down until the drag equals the weight, at which point it keeps moving at constant speed (so-called "terminal velocity"). In the hypothetical scenario where you have a constant force that's greater than the weight pulling on the object, then the object will indeed start falling upwards.

share|cite|improve this answer

edited 39 mins ago

answered 1 hour ago

AllureAllure

2,351926

$endgroup$

share|cite|improve this answer

edited 39 mins ago

answered 1 hour ago

AllureAllure

2,351926

answered 1 hour ago

AllureAllure

2,351926

answered 1 hour ago

AllureAllure

2,351926

1

$begingroup$

The acceleration (or deceleration) of an object is $a$: $$a = f/m.$$

Acceleration (or deceleration) is the rate of change of velocity. So, to find the rate of change of velocity of an object you divide all the forces acting on it, by the object's mass.

The forces acting on your hypothetical object are the object's weight and the atmospheric drag. Someone being very picky might also say that there is a relatively very small atmospheric buoyancy at work too, but it can be ignored in most cases.

Let's say the object is falling straight down. Atmospheric drag is velocity-dependent, with low drag at low speeds and much higher drag at higher speeds. Drag will cause the object to decelerate, until drag equals the object's weight. At that point the object is said to be moving at terminal velocity, and it just keeps falling at terminal velocity.

The whole scenario gets more complicated when altitude-dependent atmospheric density is taken into account, but what's described above captures the essence of an answer to your question.

share|cite|improve this answer

answered 50 mins ago

S. McGrewS. McGrew

9,81421239

$endgroup$

add a comment | 

1

$begingroup$

The acceleration (or deceleration) of an object is $a$: $$a = f/m.$$

Acceleration (or deceleration) is the rate of change of velocity. So, to find the rate of change of velocity of an object you divide all the forces acting on it, by the object's mass.

The forces acting on your hypothetical object are the object's weight and the atmospheric drag. Someone being very picky might also say that there is a relatively very small atmospheric buoyancy at work too, but it can be ignored in most cases.

Let's say the object is falling straight down. Atmospheric drag is velocity-dependent, with low drag at low speeds and much higher drag at higher speeds. Drag will cause the object to decelerate, until drag equals the object's weight. At that point the object is said to be moving at terminal velocity, and it just keeps falling at terminal velocity.

The whole scenario gets more complicated when altitude-dependent atmospheric density is taken into account, but what's described above captures the essence of an answer to your question.

share|cite|improve this answer

answered 50 mins ago

S. McGrewS. McGrew

9,81421239

$endgroup$

add a comment | 

$begingroup$

The acceleration (or deceleration) of an object is $a$: $$a = f/m.$$

Acceleration (or deceleration) is the rate of change of velocity. So, to find the rate of change of velocity of an object you divide all the forces acting on it, by the object's mass.

The forces acting on your hypothetical object are the object's weight and the atmospheric drag. Someone being very picky might also say that there is a relatively very small atmospheric buoyancy at work too, but it can be ignored in most cases.

Let's say the object is falling straight down. Atmospheric drag is velocity-dependent, with low drag at low speeds and much higher drag at higher speeds. Drag will cause the object to decelerate, until drag equals the object's weight. At that point the object is said to be moving at terminal velocity, and it just keeps falling at terminal velocity.

The whole scenario gets more complicated when altitude-dependent atmospheric density is taken into account, but what's described above captures the essence of an answer to your question.

share|cite|improve this answer

answered 50 mins ago

S. McGrewS. McGrew

9,81421239

$endgroup$

share|cite|improve this answer

answered 50 mins ago

S. McGrewS. McGrew

9,81421239

answered 50 mins ago

S. McGrewS. McGrew

9,81421239

answered 50 mins ago

S. McGrewS. McGrew

9,81421239