Question

The Earth exerts a gravitational force of F(r) = 2.99x10¹⁶/r² newtons (N) on an object with a mass of 75 kg located r meters 2 from the center of the Earth. Find the rate of change of force with respect to distance r at the surface of the Earth. Assume that the radius of the Earth is 6.77 x 10⁶ meters.

(Give your answer to six decimal places.)
F'(6.77 x 10⁶) =_________N/m

Answer 1

The rate of change of force with respect to distance r at the surface of the Earth is approximately -9.83 N/m.

Step-by-step explanation:

To find the rate of change of force with respect to distance r at the surface of the Earth, we need to find the derivative of the force function F(r) = 2.99x10¹⁶/r² with respect to r.

To do this, we can use the power rule for differentiation. The power rule states that the derivative of x^n with respect to x is nx^(n-1). Applying this rule to our function F(r), we get F'(r) = -2*2.99x10¹⁶/r³.

Since we are interested in finding the rate of change at the surface of the Earth, we can substitute r with the radius of the Earth: 6.77x10⁶ meters. Plugging in this value, we get F'(6.77x10⁶) = -2*2.99x10¹⁶/(6.77x10⁶)³. Calculating this, we find that F'(6.77x10⁶) ≈ -9.83 N/m.