Orbital reentry from 17000mph looks like dropping out of hyperspace [video]

Re:Orbital reentry from 17000mph looks like dropping out of ... by aj7 on Wednesday February 28, 2024 @08:59PM (#39545772)

Huh? Did you flunk freshman physics?

E=K+U=GmM/2r−GmM/r=−GmME/r. We can see that the total energy is negative, with the same magnitude as the kinetic energy. For circular orbits, the magnitude of the kinetic energy is exactly one-half the magnitude of the potential energy.

Do the calculation over.

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Re:Orbital reentry from 17000mph looks like dropping out of ... by versteegen on Thursday February 29, 2024 @01:42AM (#39546938)

You are correct (aside from the s/ME/M/ typo), but the snark isn't needed.

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Re:Orbital reentry from 17000mph looks like dropping out of ... by peeters on Thursday February 29, 2024 @09:51AM (#39550032)

No, they are not correct either.

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Re:Orbital reentry from 17000mph looks like dropping out of ... by peeters on Thursday February 29, 2024 @09:50AM (#39550022)

You're quoting this article, but you're misunderstanding it:

https://phys.libretexts.org/Bookshelves/University_Physics/U....

This is calculating the gravitational potential between two point masses. In other words, the potential energy if both masses were singular points at a distance from each other.

I was quoting the potential energy differential between being 100km over the ground, and on the ground. This is 98 MJ for a 100kg object, and that's what you have to cancel out to land back on Earth. The total gravitational energy of that object to the Earth's centre is more like 6300 MJ, but that's a meaningless number.

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