Posted by Sierra on Sunday, November 18, 2012 at 11:22pm.
Use the following numbers for this question.
MEarth 5.98e+24 kg
MJupiter 1.9e+27 kg
MSun 2e+30 kg
EarthSun distance 150000000000 m
JupiterSun distance 778000000000 m
rE 6380000 m
rJ 69000000 m
G = 6.67e11 Nm2kg2
the acceleration due to gravity at the surface of Jupiter has been found to be 26.6 m/s^2
How fast would a steel sphere be going if you let it drop 20 m from rest, near the surface of Jupiter?

Physics ASAP  drwls, Monday, November 19, 2012 at 3:15am
sqrt(2*g'*Y), where g' is the acceleration of gravity on Jupiter and Y = 20 m.
g' = g*(Mjupiter/Mearth)*(rE/rJ)^2
g = 9.8 m/s^2 (the value at earth surface)
g' = 2.72*9.81 = 26.6 m/s^2
The numbers that involve the sun are not needed to do this. I did not use G because the value of g on earth could be used.
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