The following expression for the acceleration due to gravity works well for obje

Question

The following expression for the acceleration due to gravity works well for objects near the Earth's surface (G is the gravitational constant, Me is the mass of the Earth, and Re, is the radius of the Earth): g = GMe/ Use the above equation to calculate the gravitational acceleration at an altitude of 100,000 meters above the Earth. By what percentage is this acceleration different from that on the Earth's surface? Using the universal law of gravitation, show that the gravitational acceleration experienced by an object is independent of its mass.

Explanation / Answer

Gravitational acceleration at the earth's surface= GM/(6400000)^2

Gravitational acceleration at 100000= 6.67 x 10^-11 x 5.97 x 10^ 24/ (6400000+ 100000)^2= 9.42m/s^2

At the surface, g= GM/6400000^2 = 9.72 m/s^2

Difference in fraction= 9.72-9.42/9.72= 0.0308

Gravitation law o attraction says that the force of gravitation experienced by an object is given by GMm/r^2

HEnc eacceleration is a= F/m=GM/r^2 which is independent of mass of the object

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