Ask Question
10 June, 06:54

A satellite circles the moon at a distance above its surface equal to 2 times the radius of the moon. The acceleration due to gravity of the satellite, as compared to the acceleration due to gravity on the surface of the moon is

+5
Answers (2)
  1. 10 June, 07:02
    0
    The orbit is a circle whose radius is 3 times the radius of the surface

    (both measured from the center of the moon). So the acceleration due

    to gravity at the orbital altitude is

    1/3² = 1/9 = 11.1% of its value on the surface.
  2. 10 June, 07:12
    0
    From Newton's Law of Universal Gravitation:

    F = GMm / r^2. (When at the surface).

    G = Universal gravitational constant, G = 6.67 * 10 ^ - 11 Nm^2 / kg^2.

    M = Mass of Moon

    m = Mass of Satellite

    r = distance apart, between centers = in this case it is the distance from Moon to the Satellite.

    Recall: F = mg.

    mg = GMm / r^2

    g = GM / r^2 ... (i). When at surface.

    Note when the satellite is at a distance 2 times the radius of the moon.

    Therefore, the distance between centers = 2r + r = 3r.

    Note, when need to add radius of the moon, because we are measuring from center of the satellite to center of the moon.

    From (i)

    g = GM / (3r) ^2. The distance r is replaced with 3r

    g = GM / 9r^2 = (1/9) * GM / r^2

    Therefore gravity on the satellite is (1/9) times that on the Moon.
Know the Answer?
Not Sure About the Answer?
Find an answer to your question 👍 “A satellite circles the moon at a distance above its surface equal to 2 times the radius of the moon. The acceleration due to gravity of ...” in 📗 Physics if the answers seem to be not correct or there’s no answer. Try a smart search to find answers to similar questions.
Search for Other Answers