What is the formula I would use to solve this?

Question

A hypothetical planet has seven times the mass of the earth and twice the radius of the Earth. The magnitude of the gravitational acceleration at the surface of this planet is most nearly______.

Thank you.

Answer 1

Gravitational acceleration g=GM/R^2, G is gravitational constant, R =planet's radius and M=planet's mass.
If Mass of this planet is 7 times and radius is twice, g will be
7/4 =1.75 times that of earth (=9.8m/s^2 * 1.75=17.15m/s^2)

Answer 2

For our planet: g=GM/r² [eq.1]

For the hypothetical planet:
a=G(7M)/(2r)² = 7GM/4r² = 7/4 GM/r²

Using eq.1, this becomes a=7/4 g and that's your final answer.

In other words, the gravitational acceleration at the surface of this planet is 1.75 times stronger than that at the surface of our Earth.

Answer 3

You've been given the formulas; all you need to do is examine them to see that the value of p (not g, that's for earth) is directly proportional to mass and inversely proportional to r².

In your problem, compared to earth, the mass factor is 7 and the radius factor is 2².

So, p = g*(7/4) = 17.15 m/s²

Answer 4

a = G M /r^2

Acceleration is equal to the gravitational constant times mass of the planet divided by planet's radius squared.

Answer 5

g = G * M/r^2

where

G is gravitational constant = 6.67 × 10−11 N m^2 kg^-2
M is mass in kg
r is radius in meters

Since you know g for earth = 9.8m/s^2
you can calculate earth's mass and use this in computing g for this other planet.
_______
Actually you can say that

9.8 = G*Me/Re^2

g = G*7Me/(2Re)^2 = 7/4 * G*Me/Re^2 = 7/4 * 9.8