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

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 -
**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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