Question

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 gravitational acceleration on the surface of the earth?

Answer 1

The planet has one-fifth of the gravitational acceleration of the Earth.

Step-by-step explanation:

Newton's law of gravitational attraction between two bodies of masses
`m_1` and
`m_2` separated by a distance d gives

`F = G(m_1m_2)/(d^2)` (G is a universal constant)

For any body of mass m on the Earth surface,

`F = G(Mm)/(R^2)`

Here, M and R are the mass and the radius of the Earth, respectively.

But this force is the gravitational force on the body.

`F = mg = G(Mm)/(R^2)`

`g = G(M)/(R^2)`

For a planet with 5 times the mass of Earth and 5 times its radius,

`g_p = G(5M)/((5R)^2) = G(M)/(5R^2) = (1)/(5)G(M)/(R^2) = (g)/(5)`

Hence the planet has one-fifth of the gravitational acceleration of the Earth.

Answer 2

The gravitational acceleration on earth is 5 times that on Driff.

Step-by-step explanation:

The gravitational acceleration on earth is given as:

ge = Gm/r²

Where m = mass of earth, r = radius of earth.

The gravitational acceleration on Driff is given as:

gd = GM/R²

Where M = Mass of Driff, R = radius of Driff.

Since we're told that M = 5m and R = 5r:

gd = G*5m/(5r)²

gd = Gm/5r²

Comparing this to the gravitational acceleration on earth:

ge : gd = Gm/r² : Gm/5r²

ge:gd = 5:1

The gravitational acceleration on earth is 5 times that on Driff.