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When a falling meteor is at a distance above the Earth's surface of 3.46 times the Earth's radius, what is its free-fall acceleration caused by the gravitational force exerted on it?? m/s2 toward the earth.

Question

When a falling meteor is at a distance above the Earth's surface of 3.46 times the Earth's radius, what is its free-fall acceleration caused by the gravitational force exerted on it?

? m/s2 toward the earth.

check_circleAnswer
Step 1
h 3.46R
Under free fall, acceleration is given by
a=g
2
(R+h)
9.8m/SR+3.46R)
=9.8m/s' (0.050)
- 0.492 m/s2
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h 3.46R Under free fall, acceleration is given by a=g 2 (R+h) 9.8m/SR+3.46R) =9.8m/s' (0.050) - 0.492 m/s2

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