HELP! Physics?
The acceleration of gravity upon Mars' surface is 3.75m/s/s. At a location of 'R' above Mars' surface (where 'R' is the radius of the Mars), the acceleration of gravity is closest to ____ m/s/s
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answers:
oubaas say: a = M*G/R^2
a' = M*G/(2R)^2 = M*G/(4*R^2) = a/4
since a = 3.75 = 3+3/4 = 15/4, then (15/4)/4 = 15/16 of m/sec^2 (0.93750)
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electron1 say: In this problem, the object’s weight is equal to the Universal gravitational force.

Weight = m * g
Universal gravitational force = G * M * m ÷ d^2
m * g = G * M * m ÷ d^2
g = G * M ÷ d^2

G and M are constant. So, the value of g is inversely proportional to the square of the distance from the center of Mars to the object. In this problem, the new distance is twice the original distance. So the new value of g will be one fourth of its original value.

g = ¼ * 3.75 = 0.9375 m/s^2

I hope this is helpful for you.
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