Asked by Anonymous
9. The surface of the earth is approximately 6,400 km from its center. If the mass of the earth is 6.0 x 1024 kg, what is the acceleration due to gravity near the surface?
Answers
Answered by
tchrwill
The acceleration due to gravity derives from g = µ/R^2 where
G = the acceleration due to gravity
µ = the earths gravitational constant = GM
G = the universal gravitational constant = 6.67259x10^-11
M = the mass of the earth = 5.97424x10^24
R = the mean surface radius of the earth = 6378km
Therefore, g = 6.67259x10-11(5.97424x10^24)/6,378,000^2
……………..= 9.7995m/sec^2
G = the acceleration due to gravity
µ = the earths gravitational constant = GM
G = the universal gravitational constant = 6.67259x10^-11
M = the mass of the earth = 5.97424x10^24
R = the mean surface radius of the earth = 6378km
Therefore, g = 6.67259x10-11(5.97424x10^24)/6,378,000^2
……………..= 9.7995m/sec^2
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