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16 April, 09:15

A satellite of mass 6000 kg orbits the earth and has a period of 5700 s. determine the radius of its circular orbit.

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  1. 16 April, 10:23
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    1) The gravitational force F between the earth and the satellite is given by:

    F = GMm / R²

    where

    G is the universal gravitational constant { = 6.67 x 10^-11 Nm²/kg² };

    M is the mass of the Earth { = 5.97 x 10^24 kg };

    m is the mass of the orbiting satellite;

    R is the distance between their centres of mass.

    Force F provides the centripetal force needed to keep the satellite in a circular orbit.

    Centripetal force = mω²R

    where ω is the angular velocity. So we can write:

    GMm / R² = mω²R

    {m cancels out. Divide both sides by ω² and multiply by R² }

    GM / ω² = R³

    We are given that the satellite completes one orbit ( = 2π radians) in 5700 seconds. So:

    ω = (2π / 5700) = 1.10 x 10^-3 rad/s

    R³ = (6.67 x 10^-11 * 5.97 x 10^24 / (1.10 x 10^-3) ²) = 3.28 x 10^20

    R = 6.89 x 10^6 m

    2) Gravitational force = GMm / R²

    = 6.67 x 10^-11 * 5.97 x 10^24 * 6000 / (6.89 x 10^6) ²

    = 5.03 x 10^4 newtons

    3) Altitude = radius of orbit - radius of Earth

    = 6.89 x 10^6 - 6.37 x 10^6

    = 5.2 x 10^5 m
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