Question

A planet with mass M and radius R has gravitational field g at the surface of the planet. In terms of g, what is the gravitational field at the surface of a planet with mass 3M and radius 3R?

Answer 1

The gravitational field strength at the surface of a planet is directly proportional to the mass of the planet and inversely proportional to the square of its radius. If a planet has a mass 3 times greater and a radius 3 times larger than another planet, the gravitational field strength at its surface will remain the same.

Step-by-step explanation:

The gravitational field strength at the surface of a planet is directly proportional to the mass of the planet and inversely proportional to the square of its radius. Therefore, if a planet has a mass 3 times greater and a radius 3 times larger than another planet, the gravitational field strength at its surface will remain the same.

Mathematically, we can express this as:

g = G(M/R^2)

where g is the gravitational field strength, G is the gravitational constant, M is the mass of the planet, and R is the radius of the planet.

Since both the mass and radius are multiplied by the same factor (3) for the second planet, the ratio of the gravitational field strengths for the two planets will be:

g2/g1 = (M2/R2^2)/(M1/R1^2) = (3M/9R^2)/(M/R^2) = 3/1 = 3

Therefore, the gravitational field strength at the surface of the second planet will be 3 times the gravitational field strength at the surface of the first planet.