In another solar system is planet Driff, which

has 5 times the mass of the earth and also 5

times the radius.

How does the gravitational acceleration on

the surface of Driff compare to the gravita-

tional acceleration on the surface of the earth?

1. It's 15th as much.

2. There is no gravity on Driff because 5 times 4000 miles (the radius of the earth), is 20000 miles, far beyond the pull of gravity.

3. It's 125th as much.

4. It's 25 times as great.

5. It's 5 times as much.

6. It's the same, 10 m/s2

.

It is (1/5) th as much.

Explanation:

If we apply the equation

F = G*m*M / r²

where

m = mass of a man

M₀ = mass of the planet Driff

M = mass of the Earth

r₀ = radius of the planet Driff

r = radius of the Earth

G = The gravitational constant

F = The gravitational force on the Earth

F₀ = The gravitational force on the planet Driff

g = the gravitational acceleration on the surface of the earth

g₀ = the gravitational acceleration on the surface of the planet Driff

we have

F₀ = G*m*M₀ / r₀² = G*m * (5*M) / (5*r) ²

⇒ F₀ = G*m*M / (5*r²) = (1/5) * F

If

F₀ = (1/5) * F

then

W₀ = (1/5) * W ⇒ m*g₀ = (1/5) * m*g ⇒ g₀ = (1/5) * g

It is (1/5) th as much.