 # How long would it take for an object to fall from very large distances (~300,000km)

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So I re-watched the most recent Star-Trek movie, and it bugged me that I couldn't figure out how long it would have taken for the ship to fall that far assuming zero relative motion at the start. (of all the things to be frustrated about that movie, this is what bothers me the most for some reason)

Assuming the gravitational field of earth was stronger than the moons at their initial altitude, how do you model such a scenario? I'm more interested in treating it as a two-body equation (ignoring the moon and atmospheric drag) instead of a three.

I believe I'm correct in assuming that the standard equation for freefall would not apply since GM/r^2. I have a feeling that I would have to dust off my Lagrangian mechanics and differential equation books to figure this out, and I really didn't do well in those classes. It's been 4 years since a college level physics class, I'm slipping!

I did find this page in my search for an answer: http://physics.stackexchange.com/questions/19813/calculating-gravity-when-taking-into-account-the-change-of-gravitational-force but plugging in the values returned something on the order of milliseconds, and I was picturing hours or days.

Thank you for any help you can give me

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A "straight down" fall is a degenerate orbit with semi-major axis equal to the half the starting height, and semi-minor axis zero. So all you need to do is calculate the period of an orbit with a=150000 km.

Punching the numbers into the orbital period formula, T = 2Ãâ‚¬Ã¢Ë†Å¡(a3/ÃŽÂ¼), I make T = 578160 s, or 6.7 days.

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I compared that situation to the Apollo missions, which basically "fell" back to Earth after leaving the Moon. Those return trips usually took 3-4 days.

The really weird part about Star Trek Into Darkness was that as the ship was basically in free fall to Earth, the crew and objects inside still felt acceleration by falling inside the ship, as if the ship was accelerating AWAY from Earth.

In reality, they would have felt weightlessness and floated around the ship, similar to the Apollo missions.

If we got into even more detail, weren't they closer to the Moon's SOI, causing them to crash on the Moon?

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I compared that situation to the Apollo missions, which basically "fell" back to Earth after leaving the Moon. Those return trips usually took 3-4 days.

The really weird part about Star Trek Into Darkness was that as the ship was basically in free fall to Earth, the crew and objects inside still felt acceleration by falling inside the ship, as if the ship was accelerating AWAY from Earth.

In reality, they would have felt weightlessness and floated around the ship, similar to the Apollo missions.

If we got into even more detail, weren't they closer to the Moon's SOI, causing them to crash on the Moon?

Yeah, like I said I was odd to pick that part out as the one that bugged me the most. The ship should have fallen towards the moon unless it moved during the battle. Even if the CoM was more towards earth, counting the moon's gravity would really slow down the acceleration during the fall I would think.

A "straight down" fall is a degenerate orbit with semi-major axis equal to the half the starting height, and semi-minor axis zero. So all you need to do is calculate the period of an orbit with a=150000 km.

Punching the numbers into the orbital period formula, T = 2Ãâ‚¬Ã¢Ë†Å¡(a3/ÃŽÂ¼), I make T = 578160 s, or 6.7 days.

So much simpler than I thought it would be; I didn't even think to use orbital equations for it. One question though is how does the kepler orbital period formula account for the changing gravitational acceleration? It seems like it would only take the value once: at 'a=150000'

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If you really want to be accurate with this, you'd need to calculate air resistance on the ship. (kinda hard, since we don't have one)

Ignoring that, and if the ship has no motion parallel to the planet's surface (anything resembling an orbit) wouldn't Newton be accurate enough for calculating it?

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Straight down falls... Assuming it's coming from a stationary site wrt Earth as a point (but not to Earth's rotation)

Acceleration = Ã¢Ë†Å¡((G.MÃ¢Å â€¢)/((h+RÃ¢Å â€¢)^2))

time = Ã¢Ë†Å¡((2h)/g)

But it's not counting into effect that the acceleration grows larger as you move closer... and also not counting into effect that at a point, the object will have to be slowed down by a fluid (that is our atmosphere).

Another good counting is through the use of Kepler laws, but you can't assume a value of e >= 1 (straight-down fall) I guess. Also, you stop at the distance from one of the orbit's foci at RÃ¢Å â€¢, not at 0.

Edited by YNM

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So much simpler than I thought it would be; I didn't even think to use orbital equations for it. One question though is how does the kepler orbital period formula account for the changing gravitational acceleration? It seems like it would only take the value once: at 'a=150000'
The orbital period formula is valid whether the orbit is circular or elliptical, and an object in an elliptical orbit will of course experience changing gravitational acceleration. IIRC Kepler's laws and the orbital period formular can be derived from Newton's laws of motion and gravitation.

I assumed, of course, that both the Earth's own radius and any atmospheric drag could be neglected. The distances involved are small compared to the total fall, the speeds are high meaning the times involved are yet smaller, and the two errors have opposite sign, so I think this is justified.

I also neglected the gravity of the Moon. The moon's sphere of influence is about 60,000 km in radius, so if the ship started 300,000 km from Earth then unless the Moon was near perigee the ship would be comfortable outside its SOI.

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The orbital period formula is valid whether the orbit is circular or elliptical, and an object in an elliptical orbit will of course experience changing gravitational acceleration. IIRC Kepler's laws and the orbital period formular can be derived from Newton's laws of motion and gravitation.

The problem is, you have a parabolic orbit, because if

a = 150000 km

Ap = 300000 km

Pe = 0 km

Then the value of e = 1.

Ap = a*(1+e) = 150000*(1+1) = 300000 km

Pe = a*(1-e) = 150000*(1-1) = 0 km

Edited by YNM

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The problem is, you have a parabolic orbit, because if

a = 150000 km

Ap = 300000 km

Pe = 0 km

Then the value of e = 1.

Ap = a*(1+e) = 150000*(1+1) = 300000 km

Pe = a*(1-e) = 150000*(1-1) = 0 km

It depends on how you interpret orbital eccentricity. A eccentricity of one could be a parabola. Or it could be a straight line segment, which is the case here.

Think of a circle with eccentricity zero and increasing to one. If you keep b constant and increase a you get a parabola. If you keep a constant and decrease b you have a line.

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Pretty much. The "orbit" is clearly not a parabola or a hyperbola, because those are both open trajectores while ours is closed. I'm not sure eccentricity is properly defined for the degenerate ellipse, but in any case it's not needed to get a decent approximation of the total fall time.

If you want to be more rigorous, you could consider an ellipse with some small semi-minor axis, then let that semi-minor axis go to zero.

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It depends on how you interpret orbital eccentricity. A eccentricity of one could be a parabola. Or it could be a straight line segment, which is the case here.

Think of a circle with eccentricity zero and increasing to one. If you keep b constant and increase a you get a parabola. If you keep a constant and decrease b you have a line.

Pretty much. The "orbit" is clearly not a parabola or a hyperbola, because those are both open trajectores while ours is closed. I'm not sure eccentricity is properly defined for the degenerate ellipse, but in any case it's not needed to get a decent approximation of the total fall time.

If you want to be more rigorous, you could consider an ellipse with some small semi-minor axis, then let that semi-minor axis go to zero.

e = c/a ; and for every ellipse, a^2 = b^2 + c^2. In which case that e = 1 so a = c, b will always be zero... wait a minute...

Yeah, just say that we have no tangential velocity, so a zero b is a must. Then, for a 300000 km fall to Earth's center, your fall time is less than 6d 16h 16m 43.19s , assuming that G = 6.67*10^(-11) m^3 kg^(-2) s^(-2) and MÃ¢Å â€¢ = 6*10^(24) kg.

Edited by YNM

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