A newly discovered planet has a radius twice as large as earth's and a mass five times as large. What is the free-fall acceleration on its surface?

free-fall acceleration = 12.25 m/s²

Explanation:

Formula for free fall acceleration is given by the gravity equation;

g = Gm/r²

Where;

M is mass of the Earth

r is radius of the Earth

G is the gravitational constant

Now, that equation applies to the earth.

For this new planet, we are told that the mass = 5 × mass of earth

And radius = 2 × radius of earth.

Thus, free fall acceleration for this planet is;

g_p = G(5m)/(2r)²

g_p = (5/4)(Gm/r²)

Gravitational constant has a value of 6.674 × 10^(−11) N.m²/kg²

Mass of the earth = 5.972 × 10^(24) kg

Radius of the earth = 6378 km = 6378000

Thus;

g_p = (5/4)(6.674 × 10^(−11) × 5.972 × 10^(24))/(6378000²) = 12.25 m/s²

i think the answer is a and b

i wish you luck