What is the acceleration due to gravity on the surface of a planet that has twice the mass of the Earth and half its radius?

1 Answer

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_1m_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_1m_2)/r^2#

then make the substitutions:

#F_g=G((2m_1)m_2)/(1/2r)^2#

#F_g=G((2m_1)m_2)/(1/4r)#

So the numerator increases linearly (#xx2# ) but the denominator reduces by an exponential - in this case (#xx4# ).

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#