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elena55 [62]
3 years ago
8

The free-fall acceleration at the surface of planet 1 is 26 m/s^2 . The radius and the mass of planet 2 are twice those of plane

t 1.
What is the free-fall acceleration on planet 2? Express your answer using two significant figures. m/s^2.
Physics
1 answer:
valkas [14]3 years ago
6 0

Answer:

13 m/s^2

Explanation:

The acceleration of gravity near the surface of a planet is:

g = MG / R^2

For planet 1, g = 26 m/s^2.

The gravity on planet 2 in terms of the mass and radius of planet 1 is:

g = (2M)G / (2R^2)

g = 1/2 MG / R^2

Since MG/R^2 = 26 m/s^2, then:

g = 13 m/s^2

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1) t=1.743 sec

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