What is the gravitational force between the Earth (mEarth = 5.98 x 1024 kg, rEarth = 6.378 x 106 m) and a 15,000 kg satellite in

Question

What is the gravitational force between the Earth
(mEarth = 5.98 x 1024 kg, rEarth = 6.378 x 106 m)
and a 15,000 kg satellite in Earth’s orbit 575-km above Earth’s surface?

Answer:
____________ Newtons
I really need help on this! Can someone please help? Thank you!

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Gia Bảo 4 years 2021-08-09T12:26:08+00:00 1 Answers 59 views 0

Answers ( )

  1. Ladonna
    0
    2021-08-09T12:27:23+00:00
    Reply

    Answer:

    17.7MN

    Explanation:

    Given Data

    m1 = 5.98 x 10^24 kg

    m2= 15,000 kg

    R= 6.378 x 10^6 m+575-km

    R= 6.378 x 10^3+575*10^3-km

    R= 581378m

    G = 6.673 x 10-11 N m^2/kg2

    This net centripetal force is the result of the gravitational force that attracts the satellite towards the central body and can be represented as

    F =GM1M2 / R^2

    F =6.673 x 10-11* 5.98 x 10^24*15,000 /  581378^2

    F =6.673 x 10^-11* 5.98 x 10^24*15,000 /  581378^2

    F= 39.90454*10^13*15,000/  581378^2

    F= 598.5681*10^16/338000378884

    F= 17709095.5335N

    F= 17.7MN

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