Question

Let us consider a mythical planet has twice the mass and twice the radius of Earth. compute the acceleration due to gravity at its surface.​

Answer 1

Answer:

The mass increases linearly but radius decreases exponentially, so the result is

9.8

m

s

2

⋅

2

⋅

4

=

78.4

m

s

2

Explanation:

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