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February 27, 2015

February 27, 2015

Posted by **Mikhail** on Thursday, April 15, 2010 at 12:43am.

What is the speed of a satellite orbiting 4.70km above the surface?

What is the escape speed from the asteroid?

- Physics -
**drwls**, Thursday, April 15, 2010 at 1:49amYou need to provide units for your quantities. I will have to assume your 8.50 is km and your 10^16 is kg.

4.70 km above the surface is R = 13.2 km (or 1.32*10^4 m) from the center

The orbital speed satisfies

G M/R^2 = V^2/R

where M is the asteroid mass and you know what G is. Solve for V.

V = sqrt(GM/R)

For the escape velocity Ve from the asteroid FROM THE SURFACE, where R' = 8.50 *10^3 m,

Ve^2/2 = GM/R'

Ve = sqrt(2 G M/R')

Note that a different radius, R', must be used.

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