what is the acceleration due to gravity on the surface of a planet that has a radius half that of earth and same avarage density as earth
Answers
Answer:
Let's first look at the equation for the force of gravity:
F
g
=
G
m
1
m
2
r
2
which is often simplified for working with objects on the surface of the Earth (since we know the gravitational constant and the mass of the Earth) to
F
g
=
M
r
2
where M is the mass experiencing Earth's gravity.
So what happens when we double the mass of the Earth and reduce its radius to 1/2? Let's multiply
m
1
by 2 and substitute in
1
2
r
for
r
. So first start with the full equation:
F
g
=
G
m
1
m
2
r
2
then make the substitutions:
F
g
=
G
(
2
m
1
)
m
2
(
1
2
r
)
2
F
g
=
G
(
2
m
1
)
m
2
1
4
r
So the numerator increases linearly (
×
2
) but the denominator reduces by an exponential - in this case (
×
4
).
The force of gravity on Earth is roughly
9.8
m
s
2
but on this other planet, it would be:
9.8
m
s
2
⋅
2
⋅
4
=
78.4
m
s
2