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Apollo 13 - A Puzzle

Sun Apr 08, 2001 1:07 pm

I just watched Apollo 13 on tv tonight, and I noticed something that confused me.

When preparing for reentry, Mission Control informs the crew that their trajectory is still a bit shallow. This is because of the lighter weight of the craft due to not having any lunar cargo. They are told to ballast the craft by bringing items of the LEM into the Control Module (is that the correct term?)...thereby making it heavier.

But they're in space! There is no gravity. How can this possibly work? Could this be an oversight by the film's producers?
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RE: Apollo 13 - A Puzzle

Sun Apr 08, 2001 1:16 pm

Prepairing for reentry is reentry into the atmosphere, therefore if they bring those items into the controll room when they enter the Earth's atmosphere they will have more weight.
I'd like to elect a president that has a Higher IQ than a retarted ant.
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RE: Apollo 13 - A Puzzle

Sun Apr 08, 2001 1:37 pm

There technically is gravity in space. The spacecraft is being affected by the Earth's gravity. So the MASS (not weight) is critical in getting the right entry angle.
Stop drop and roll will not save you in hell. --- seen on a church marque in rural Virginia
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RE: Apollo 13 - A Puzzle

Sun Apr 08, 2001 10:25 pm

When they're that close to the Earth, they are affected by Earth's gravity. It's the same principle as how satellites are kept in orbit. Earth's gravity.
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RE: Apollo 13 - A Puzzle

Sun Apr 08, 2001 11:25 pm

Good morning,

On the Apollo missions it was expected that they would be carrying back a coulble hundred pounds of lunar rocks with them. By assuming that they would have rocks, they could account that in the ships trajectory, thus setting the ship up for re-entry a bit easier. On Apollo 13, they did not have this extra weight, and as a result, there trajectory was a little off. By ballasting items from the LEM, they can add that weight in to the Command Module thus making it heavier for re-entry. If they did not do that, there would be the chance that they would hit the earth at a slighty different angle, and go skipping away from earth.

Now what about on previos missions such as Apollo 8? Well, it was not calcualted in that they were suppose to carry moon rocks, because they did not land on the moon. So as a result, the numbers were all different, and they would not come up "shallow"

I hope that answers your question.
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RE: Apollo 13 - A Puzzle

Mon Apr 09, 2001 6:29 pm

Hi there,

I'll give you the tech stuff involved in your question.

Newton's law of gravity says:

F= G x M1 x M2 / r^2

Which means the following:

The (gravitational) force between two bodies is equal to G multiplied by M1, multiplied by M2, and all this devided by r to the power of 2.


G = the gravity constant (6.672 x 10^-11)
M1 = the mass of the first body
M2 = the mass of the second body
r = the distance between the 2 objects

The gravity constant is a universal constant and is effective throughout the whole universe.

Now you can see that the gravitational force is inversely proportional to the distance between the two objects. This means that the gravitational force only approached zero when the distance becomes infinite, which off course doesn't exist, the distance can only become very great (in terms of light years) but never infinite.

So for every possible distance between the two objects, there will always be a gravitational force between them.

BTW: this principle is used when sending space probes to other planets. If they had to use a rocket motor all the way, the rocket would become too heavy (too expensive and maybe technically not achievable). So they wait until the planets align properly and then they launch the space probe (for example the 2 Voyager probes) so that they will go from one planet to another, each time picking up some speed from the gravitational attraction of the planets. Its is commonly reffered to as playing "cosmic billiards".

Also, although being 150 million kilometers away from us, the sun has definately a gravitational effect on us as she is responsible (in coöperation with the moon) for the tides of the sea.

So, I hope this answers your question.

Best regards,

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