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A m = 88.1 kg object is released from rest at a distance h = 1.08246 R above the Earth’s surface.
The acceleration of gravity is 9.8 m/s2 .For the Earth, RE = 6.38×106 m, M = 5.98 × 1024 kg. The gravitational acceleration at the surface of the earth is g = 9.8 m/s2.
Find the speed of the object when it strikes the Earth’s surface. Neglect any atmospheric friction.
Caution: You must take into account that the gravitational acceleration depends on dis- tance between the object and the center of the earth.
Answer in units of m/s.
I am pretty lost on this problem.
Does the solution solve to be v^2(f)= (2GM)/r ? if so, HOW?
 
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The post A m = 88.1 kg object is released from rest at a distance h = 1.08246 R above the Earth’s surface. The acceleration of gravity is 9.8 m/s2 .For the Earth, RE = 6.38×106 m, M = 5.98 × 1024 kg. appeared first on Superb Professors.

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