Physics

When the space shuttle (mass of the shuttle = 2 x10^6 kg) was at the outer limit of Earth's atmosphere (distance = 350 km above Earth's surface), astronauts measured the gravitational force from Earth to be equal to 1.7 x 10^7 newtons. Assuming that both Earth and the shuttle are point masses, calculate the gravitational force of Earth on the shuttle when the shuttle was at the limit of the mesosphere, i.e., at 90 km from Earth's surface. Also, what is the gravitational force the shuttle exerts on Earth at this same point?

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asked by Leila
  1. F = G Mearth Mshuttle /r^2
    G , Mearth and Mshuttle are the same

    so
    F r^2 = same
    F1 r1^2 = F2 r2^2
    F2 = F1 r1^2/r2^2

    F1 = 1.7*10^7
    r1 = 350*10^3 + 6.38*10^6 = 6.73*10^6 meters
    r2 = 90*10^3 + 6.38*10^6 = 6.47*10^6 meters
    so
    F2 = 1.7*10^7 (6.73/6.47)^2
    = 1.7 *10^7 * 1.082
    = 1.84 * 10^7 Newtons

    Newton's third law applies o the second part of the question.

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    posted by Damon

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