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1- If a satellite orbiting the Earth at a height of 150 km (93 miles) above the surface of the earth.
Calculate the speed, acceleration and orbital period of the satellite. Using these given constants:
M(earth) = 5.98 x 10^24 kg, R(earth) = 6.37 x 10^6 m).

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Anonymous

  1. F = G m M/r^2
    G = 6.67*10^-11
    m is mass of our thing
    M = 5.98*10^24
    r = 6.37*10^6 + 150,000 = 6.37*10^6 + 0.150 *10^6 = 6.52*10^6
    so
    F = m * 6.67*10^-11 * 5.98*10^24 / 42.5*10^12
    = m * 9.38
    {{ note - if we were at earth surface it would be about m * 9.81 , we did not go up much :) }}
    Now the problem:
    F = mass * centripetal a = m * v^2/r
    so
    9.38 = v^2/r = v^2 / 6.52*10^6
    v^2 = 9.38 * 6.52 * 10^6 meters/second
    solve for v
    then a = v^2/r
    T = 2 pi r / v

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    Damon

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