A 213.7 kg satellite is in a circular orbit of 22,236 miles (35,768 km) in radius. The force keeping the satellite in orbit is 44.1 N. What

Question

A 213.7 kg satellite is in a circular orbit of 22,236 miles (35,768 km) in radius. The force keeping the satellite in orbit is 44.1 N. What is the velocity (speed) of the satellite

in progress 0
Khoii Minh 4 years 2021-08-28T10:33:11+00:00 2 Answers 66 views 0

Answers ( )

  1. Latifah
    0
    2021-08-28T10:34:32+00:00
    Reply

    Answer:

    2716.84 m/s

    Explanation:

    Using,

    F = mv²/r…………….. Equation 1

    Where F = The force keeping the satellite in the orbit, m = mass of the satellite, v = velocity of the satellite, r = radius of the circular orbit.

    make v the subject of equation 1

    v = √(Fr/m)……………… Equation 2

    Given: F = 44.1 N, r = 35768 km = 35768000 m, m = 213.7 kg.

    Substitute into equation 2

    v = √(44.1×35768000/213.7)

    v = √(7381229.761)

    v = 2716.84 m/s

    Hence the velocity of the satellite = 2716.84 m/s

  2. RuslanHeatt
    0
    2021-08-28T10:34:49+00:00
    Reply

    Answer:

    v = 3.08 k ms^-1

    Explanation:

    As we know that, v=\sqrt{GM/r}

    where r = R + h = 35768 + 6400 = 42168 km = 4.2168 x 10^7 m

    G = 6.673 x 10^-11 N m^-2 kg^-2 and M = 6.0 x 10^24 kg

    By substituting the required values, we get…

    v = \sqrt{}6.673 x 10^-11 x 6.0 x 10^24 / 4.2168 x 10^7 = \sqrt{}9.494 x 10^6 ms^-1 = 3081.2 ms^-1 = 3.08 k ms^-1

Leave an answer

Browse

Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )