# 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?

The mass increases linearly but radius decreases exponentially, so the result is
##9.8 m/s^2 * 2 * 4 = 78.4 m/s^2##

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##

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