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
.

Respuesta :

Answer:

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.

The gravitational acceleration on the surface of Driff compare to the gravitational acceleration on the surface of the earth should be (1/5)th as much.

Calculation of the  gravitational acceleration:

Here we applied the following equation

F = G*m*M / r²

Here

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

So,

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

F₀ = G*m*M / (5*r²) = (1/5)*F

Now

F₀ = (1/5)*F

So,

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

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