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 planet 1.

what is the free-fall acceleration on planet 2? express your answer using two significant figures. m/s^2.

The free-fall acceleration on planet 2 is "13 m/s²".

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Thus the above answer is right.

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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

the answer is f1 and d2 because if you put equal force on those two points the beam would balance trust me on this