The International Space Station ISS (assume that the weight of the ISS, if it we

Question

The International Space Station ISS (assume that the weight of the ISS, if it were on the surface of the earth, is approximately 990,000 pounds) orbits the earth at an altitude of approximately 250 miles above the surface of the earth. Assuming a circular orbit, what is the attractive force that the earth exerts to keep it in orbit? Give your answer in the form "a.bc x 10A(y)" N. Note that the unit representation should not be included in your answer input since it is not inside the "" symbols!!

Explanation / Answer

First, I will convert all the units to standard S.I units

1 pound = 0.453592 Kg

Therefore, 990000 pound = 449056.446 Kg

Similarly,

1 mile = 1609.34 m

Therefore,

250 miles = 402336 m

Force of attraction can be found by

F = GMm/r2

G is gravitational constant

where M is mass of earth = 5.98e24 Kg

m is mass of ISS = 449056.446 Kg

r is the distance from center of earth to center of ISS.

Note we are given distance above the surface, not from the center of earth, therefore, we need to add radius of earth,

r = 6371 Km + 402.336 Km

r = 6773.336 Km

Therefore,

F = 6.673e-11*5.98e24 *449056.446 / 6773.336 * 1000

F = 2.645e13 N

Note - If you do not want to include the distance from center of earth, you can do so by using the same method.

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