Posted by **Leila** on Sunday, May 20, 2012 at 4:29pm.

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?

- Physics -
**Damon**, Sunday, May 20, 2012 at 6:21pm
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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