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