Asked by Adam
An astronaut stands at a position on the Moon such that Earth is directly overhead and releases a Moon rock that was in her hand.
(c) What is the gravitational force exerted by the Earth on the same 2.7-kg rock resting on the surface of the Moon?
answer in mN
I keep calculating this but am getting it wrong please help
((6.67x10^-11Nm^2) / kg^2 x 2.7kg x (7.35x10^22kg) / (3.8226x10^8m)^2).
gives me about 9.06x10^-5 N so mN its about .09 mN but its wrong.
(c) What is the gravitational force exerted by the Earth on the same 2.7-kg rock resting on the surface of the Moon?
answer in mN
I keep calculating this but am getting it wrong please help
((6.67x10^-11Nm^2) / kg^2 x 2.7kg x (7.35x10^22kg) / (3.8226x10^8m)^2).
gives me about 9.06x10^-5 N so mN its about .09 mN but its wrong.
Answers
Answered by
Elena
F =G•m•M(E)/[R(EM) –R(M)]²,
where
the gravitational constant G =6.67•10^-11 N•m²/kg²,
m =2.7 kg,
Earth’s mass is M(E) = 5.97•10^24 kg,
Distance between the centres of the Earth
and the Moon
R(EM) =384467000 m.
Moon’ radius R(M) =1737000 m.
My result is F = 7.34•10^-3 N
where
the gravitational constant G =6.67•10^-11 N•m²/kg²,
m =2.7 kg,
Earth’s mass is M(E) = 5.97•10^24 kg,
Distance between the centres of the Earth
and the Moon
R(EM) =384467000 m.
Moon’ radius R(M) =1737000 m.
My result is F = 7.34•10^-3 N
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