# Physics ASAP

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Use the following numbers for this question.
MEarth 5.98e+24 kg
MJupiter 1.9e+27 kg
MSun 2e+30 kg
Earth-Sun distance 150000000000 m
Jupiter-Sun distance 778000000000 m
rE 6380000 m
rJ 69000000 m
G = 6.67e-11 Nm2kg-2
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 -

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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