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 %%

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Answer:

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

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%

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