The earth exerts a gravitational force of F (r) = 2.991016r2 (in Newtons) on an object with a mass of 75 kg, where r is the distance (in meters) from the center of the earth. Find the rate of change of force with respect to distance at the surface of the earth, assuming the radius of the earth is 6.77106 m.

Question

Physics

Albert Huffman

8 August, 20:28

The earth exerts a gravitational force of F (r) = 2.991016r2 (in Newtons) on an object with a mass of 75 kg, where r is the distance (in meters) from the center of the earth. Find the rate of change of force with respect to distance at the surface of the earth, assuming the radius of the earth is 6.77106 m.

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Answers (1)

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Robert Gilmore · Answer 1

-1.927 * 10^-4 N/m.

Explanation:

From Newton's law, F is proportional to 1/r^2, taking the derivative in respect to r of both sides of the equation

F = (2.99x10^16) / r^2

So you do that and you get

dF/dr = - 2 (2.99 * 10 ^16) / (r^3)

Given:

r = 6.77*10^6 m

dF/dr = - 2 (2.99 * 10 ^16) / (6.77*10^6) ^3

= - 1.927 * 10^-4 N/m.