Question

If g is the acceleration due to gravity on the surface of earth, its value at a height equal to triple

the radius of earth is (Assuming the Earth to be perfect sphere)​

Answer 1

Answer:

The radius of Earth at the equator is 3,963 miles (6,378 kilometers), according to NASA's Goddard Space Flight Center. However, Earth is not quite a sphere.

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Answer 2

Answer:

g/16

Explanation:

The trick of the trade here is to identify that at 3R distance above the earth is actually 4R distance away from the earth’s center ! Then you have an easy home run. So g = GM/R^2 on earth’s surface. Hence g’ = GM/(4R)^2 = GM/16R^2. So ratio of g’/g can be mathematically expressed as follows : g’/g = (GM/16R^2)/GM/R^2 = 1/16. Hence g’ = g/16. That is 1/16 of g on earth’s surface and near the surface. The value of g is 9.81m/s^2 hence g’ is 0.6131 m/s^2 at 3R distance above the earth’s surface. So here’s more information : The earth’s radius is 3,958.8 miles. Hence that acceleration of 0.6131 m/s^2 will occur at 11,876.4 miles above the earth’s surface which is 11,576.4 miles above the earth’s atmosphere’s upper limit of 300 miles. So it is way way above in space above the earth’s upper atmospheric limit that that g’ value will occur