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3

$begingroup$

Somewhat inspired by this question and its answers, does it require less delta-v for an object to fall into the Sun the further away it is from the Sun?

It makes sense that an object has to shed its orbital velocity before falling into the Sun, but counterintuitive that it is harder to reach the Sun from Earth (or even Mercury) than it is from Pluto.

orbital-mechanics gravity

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  $begingroup$
  To collide with the sun? Yes, it takes less delta-v from a more distant circular orbit. If you want a circular orbit close to the sun, it'll take more delta-v, though.
  $endgroup$
  – Ghedipunk
  8 hours ago
  

  
  
  
  

add a comment
 | 

3

$begingroup$

Somewhat inspired by this question and its answers, does it require less delta-v for an object to fall into the Sun the further away it is from the Sun?

It makes sense that an object has to shed its orbital velocity before falling into the Sun, but counterintuitive that it is harder to reach the Sun from Earth (or even Mercury) than it is from Pluto.

orbital-mechanics gravity

share|improve this question

asked 9 hours ago

TipTapTipTap

7644 bronze badges

New contributor

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

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  To collide with the sun? Yes, it takes less delta-v from a more distant circular orbit. If you want a circular orbit close to the sun, it'll take more delta-v, though.
  $endgroup$
  – Ghedipunk
  8 hours ago
  

  
  
  
  

add a comment
 | 

$begingroup$

Somewhat inspired by this question and its answers, does it require less delta-v for an object to fall into the Sun the further away it is from the Sun?

It makes sense that an object has to shed its orbital velocity before falling into the Sun, but counterintuitive that it is harder to reach the Sun from Earth (or even Mercury) than it is from Pluto.

orbital-mechanics gravity

share|improve this question

asked 9 hours ago

TipTapTipTap

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New contributor

TipTap 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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share|improve this question

asked 9 hours ago

TipTapTipTap

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share|improve this question

asked 9 hours ago

TipTapTipTap

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New contributor

TipTap is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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asked 9 hours ago

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6

$begingroup$

Yes.

1st scenario: A spacecraft orbiting the Sun at Earth distance vs. Pluto distance, shedding its orbital velocity

The orbital velocity decreases with distance, according to the following formula, where $r$ is the orbital radius, and $mu$ is the mass parameter (it's just a shorthand we use)

$$v_{circular} = sqrt{frac{mu}{r}}$$

The orbital velocity at Earth distance is 30 km/s, at Pluto it's 4.7 km/s. Shedding this velocity to fall straight down is clearly easier at Pluto distance.

(we don't quite need to fall straight down, we can still have some horizontal velocity as the Sun is not a point, but it doesn't change the qualitative answer)

2nd scenario: A spacecraft orbiting the Sun at Earth distance vs. Pluto distance, shedding its orbital velocity, but this time a little smarter

If falling into the Sun was easier when farther away, why don't we first try to go farther out then? Turns out that's a little bit more efficient.

You can't get farther away than escaping the Solar system. If you do that, and you "at infinity" so to speak is crawling away at basically 0 velocity, you can just make a tiny burn with the rocket engine to turn around and fall back into the Sun.

So how much does it cost to escape?

Escape velocity can be calculated in the following way:

$$v_{escape} = sqrt{frac{2mu}{r}} = sqrt{2}cdot v_{circular}$$

We are already travelling at circular velocity, so the extra velocity change needed is $v_e - v_c approx 0.41 v_c$

Crashing into the Sun from Earth distance now costs only 12 km/s, and from Pluto distance 1.9 km/s. Which one is cheaper didn't change, since we multiplied by the same constant (0.41)

3rd scenario: But what if the Earth and Pluto are still there?

If we start from the surface of (or an orbit around) these locations, the calculation has an extra step, as we have to escape their gravitational field first.

After escaping Pluto, we would want to have solar system escape velocity, and this we know is 1.9 km/s larger than the speed Pluto is travelling at. We want this at a sufficiently large distance from Pluto, so let's call it $v_{infty}$

The following equation is what we use to reach a target velocity after escaping:

$$v^2 = v_{infty}^2 + v_e^2$$

or

$$v = sqrt{v_{infty}^2 + v_e^2}$$

Escape velocity at the surface of Pluto is 1.2 km/s, so calculating the $v$ above gives us a required burn of 2.3 km/s.

2.3 km/s is not enough to even get into low Earth orbit, and it's not even enough to escape Earth when starting from orbit.

It's thus easier to reach the Sun from Pluto than Earth.

share|improve this answer

answered 7 hours ago

Hohmannfan♦Hohmannfan

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6

$begingroup$

Yes.

1st scenario: A spacecraft orbiting the Sun at Earth distance vs. Pluto distance, shedding its orbital velocity

The orbital velocity decreases with distance, according to the following formula, where $r$ is the orbital radius, and $mu$ is the mass parameter (it's just a shorthand we use)

$$v_{circular} = sqrt{frac{mu}{r}}$$

The orbital velocity at Earth distance is 30 km/s, at Pluto it's 4.7 km/s. Shedding this velocity to fall straight down is clearly easier at Pluto distance.

(we don't quite need to fall straight down, we can still have some horizontal velocity as the Sun is not a point, but it doesn't change the qualitative answer)

2nd scenario: A spacecraft orbiting the Sun at Earth distance vs. Pluto distance, shedding its orbital velocity, but this time a little smarter

If falling into the Sun was easier when farther away, why don't we first try to go farther out then? Turns out that's a little bit more efficient.

You can't get farther away than escaping the Solar system. If you do that, and you "at infinity" so to speak is crawling away at basically 0 velocity, you can just make a tiny burn with the rocket engine to turn around and fall back into the Sun.

So how much does it cost to escape?

Escape velocity can be calculated in the following way:

$$v_{escape} = sqrt{frac{2mu}{r}} = sqrt{2}cdot v_{circular}$$

We are already travelling at circular velocity, so the extra velocity change needed is $v_e - v_c approx 0.41 v_c$

Crashing into the Sun from Earth distance now costs only 12 km/s, and from Pluto distance 1.9 km/s. Which one is cheaper didn't change, since we multiplied by the same constant (0.41)

3rd scenario: But what if the Earth and Pluto are still there?

If we start from the surface of (or an orbit around) these locations, the calculation has an extra step, as we have to escape their gravitational field first.

After escaping Pluto, we would want to have solar system escape velocity, and this we know is 1.9 km/s larger than the speed Pluto is travelling at. We want this at a sufficiently large distance from Pluto, so let's call it $v_{infty}$

The following equation is what we use to reach a target velocity after escaping:

$$v^2 = v_{infty}^2 + v_e^2$$

or

$$v = sqrt{v_{infty}^2 + v_e^2}$$

Escape velocity at the surface of Pluto is 1.2 km/s, so calculating the $v$ above gives us a required burn of 2.3 km/s.

2.3 km/s is not enough to even get into low Earth orbit, and it's not even enough to escape Earth when starting from orbit.

It's thus easier to reach the Sun from Pluto than Earth.

share|improve this answer

answered 7 hours ago

Hohmannfan♦Hohmannfan

13.3k11 gold badge4848 silver badges106106 bronze badges

$endgroup$

add a comment
 | 

6

$begingroup$

Yes.

1st scenario: A spacecraft orbiting the Sun at Earth distance vs. Pluto distance, shedding its orbital velocity

The orbital velocity decreases with distance, according to the following formula, where $r$ is the orbital radius, and $mu$ is the mass parameter (it's just a shorthand we use)

$$v_{circular} = sqrt{frac{mu}{r}}$$

The orbital velocity at Earth distance is 30 km/s, at Pluto it's 4.7 km/s. Shedding this velocity to fall straight down is clearly easier at Pluto distance.

(we don't quite need to fall straight down, we can still have some horizontal velocity as the Sun is not a point, but it doesn't change the qualitative answer)

2nd scenario: A spacecraft orbiting the Sun at Earth distance vs. Pluto distance, shedding its orbital velocity, but this time a little smarter

If falling into the Sun was easier when farther away, why don't we first try to go farther out then? Turns out that's a little bit more efficient.

You can't get farther away than escaping the Solar system. If you do that, and you "at infinity" so to speak is crawling away at basically 0 velocity, you can just make a tiny burn with the rocket engine to turn around and fall back into the Sun.

So how much does it cost to escape?

Escape velocity can be calculated in the following way:

$$v_{escape} = sqrt{frac{2mu}{r}} = sqrt{2}cdot v_{circular}$$

We are already travelling at circular velocity, so the extra velocity change needed is $v_e - v_c approx 0.41 v_c$

Crashing into the Sun from Earth distance now costs only 12 km/s, and from Pluto distance 1.9 km/s. Which one is cheaper didn't change, since we multiplied by the same constant (0.41)

3rd scenario: But what if the Earth and Pluto are still there?

If we start from the surface of (or an orbit around) these locations, the calculation has an extra step, as we have to escape their gravitational field first.

After escaping Pluto, we would want to have solar system escape velocity, and this we know is 1.9 km/s larger than the speed Pluto is travelling at. We want this at a sufficiently large distance from Pluto, so let's call it $v_{infty}$

The following equation is what we use to reach a target velocity after escaping:

$$v^2 = v_{infty}^2 + v_e^2$$

or

$$v = sqrt{v_{infty}^2 + v_e^2}$$

Escape velocity at the surface of Pluto is 1.2 km/s, so calculating the $v$ above gives us a required burn of 2.3 km/s.

2.3 km/s is not enough to even get into low Earth orbit, and it's not even enough to escape Earth when starting from orbit.

It's thus easier to reach the Sun from Pluto than Earth.

share|improve this answer

answered 7 hours ago

Hohmannfan♦Hohmannfan

13.3k11 gold badge4848 silver badges106106 bronze badges

$endgroup$

add a comment
 | 

$begingroup$

Yes.

1st scenario: A spacecraft orbiting the Sun at Earth distance vs. Pluto distance, shedding its orbital velocity

The orbital velocity decreases with distance, according to the following formula, where $r$ is the orbital radius, and $mu$ is the mass parameter (it's just a shorthand we use)

$$v_{circular} = sqrt{frac{mu}{r}}$$

The orbital velocity at Earth distance is 30 km/s, at Pluto it's 4.7 km/s. Shedding this velocity to fall straight down is clearly easier at Pluto distance.

(we don't quite need to fall straight down, we can still have some horizontal velocity as the Sun is not a point, but it doesn't change the qualitative answer)

2nd scenario: A spacecraft orbiting the Sun at Earth distance vs. Pluto distance, shedding its orbital velocity, but this time a little smarter

If falling into the Sun was easier when farther away, why don't we first try to go farther out then? Turns out that's a little bit more efficient.

You can't get farther away than escaping the Solar system. If you do that, and you "at infinity" so to speak is crawling away at basically 0 velocity, you can just make a tiny burn with the rocket engine to turn around and fall back into the Sun.

So how much does it cost to escape?

Escape velocity can be calculated in the following way:

$$v_{escape} = sqrt{frac{2mu}{r}} = sqrt{2}cdot v_{circular}$$

We are already travelling at circular velocity, so the extra velocity change needed is $v_e - v_c approx 0.41 v_c$

Crashing into the Sun from Earth distance now costs only 12 km/s, and from Pluto distance 1.9 km/s. Which one is cheaper didn't change, since we multiplied by the same constant (0.41)

3rd scenario: But what if the Earth and Pluto are still there?

If we start from the surface of (or an orbit around) these locations, the calculation has an extra step, as we have to escape their gravitational field first.

After escaping Pluto, we would want to have solar system escape velocity, and this we know is 1.9 km/s larger than the speed Pluto is travelling at. We want this at a sufficiently large distance from Pluto, so let's call it $v_{infty}$

The following equation is what we use to reach a target velocity after escaping:

$$v^2 = v_{infty}^2 + v_e^2$$

or

$$v = sqrt{v_{infty}^2 + v_e^2}$$

Escape velocity at the surface of Pluto is 1.2 km/s, so calculating the $v$ above gives us a required burn of 2.3 km/s.

2.3 km/s is not enough to even get into low Earth orbit, and it's not even enough to escape Earth when starting from orbit.

It's thus easier to reach the Sun from Pluto than Earth.

share|improve this answer

answered 7 hours ago

Hohmannfan♦Hohmannfan

13.3k11 gold badge4848 silver badges106106 bronze badges

$endgroup$

share|improve this answer

answered 7 hours ago

Hohmannfan♦Hohmannfan

13.3k11 gold badge4848 silver badges106106 bronze badges

answered 7 hours ago

Hohmannfan♦Hohmannfan

13.3k11 gold badge4848 silver badges106106 bronze badges

answered 7 hours ago

Hohmannfan♦Hohmannfan

13.3k11 gold badge4848 silver badges106106 bronze badges