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
Mâ‚€ = mass of the planet Driff
M = mass of the Earth
râ‚€ = radius of the planet Driff
r = radius of the Earth
G = The gravitational constant
F = The gravitational force on the Earth
Fâ‚€ = The gravitational force on the planet Driff
g = the gravitational acceleration on the surface of the earth
gâ‚€ = the gravitational acceleration on the surface of the planet Driff
we have
Fâ‚€ = G*m*Mâ‚€ / r₀² = G*m*(5*M) / (5*r)²  Â
⇒  F₀ = G*m*M / (5*r²) = (1/5)*F
If
Fâ‚€ = (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
Mâ‚€ = mass of the planet Driff
M = mass of the Earth
râ‚€ = radius of the planet Driff
r = radius of the Earth
G = The gravitational constant
F = The gravitational force on the Earth
Fâ‚€ = The gravitational force on the planet Driff
g = the gravitational acceleration on the surface of the earth
gâ‚€ = the gravitational acceleration on the surface of the planet Driff
So,
Fâ‚€ = G*m*Mâ‚€ / r₀² = G*m*(5*M) / (5*r)²  Â
F₀ = G*m*M / (5*r²) = (1/5)*F
Now
Fâ‚€ = (1/5)*F
So,
W₀ = (1/5)*W  ⇒  m*g₀ = (1/5)*m*g  ⇒  g₀ = (1/5)*g
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