Question

A 2900 kg satellite used in a cellular telephone network is in a circular orbit at a height of 770 km above the surface of the earth. Part A What is the gravitational force on the satellite? FF = −28420 N SubmitPrevious AnswersRequest Answer Incorrect; Try Again; 18 attempts remaining Part B What fraction is this force of the satellite's weight at the surface of the earth? nothing %%

Answer 1

a) F = 2.26897 10⁴ N and b) F / W% = 79.8%

Step-by-step explanation:

For this calculation we must use the law of universal gravitation

F = G m M / r²

Where r is the distance from the center of the earth to the satellite

R = Re + y

R = 6.370 10⁶ + 770 10³ =

R = 7.140 10⁶ m

Let's calculate the strength

F = 6.67 10⁻¹¹ 2900 5.98 10²⁴ / (7.140 10⁶)²

F = 2.26897 10⁴ N

B) the fraction of the force to the weight of the satellite

W = mg

W= 2900 9.8

W = 2,842 104 N

F / W = 2.26897 10⁴ / 2.842 10⁴

F / W = 0.7984

F / W% = 79.8%