# 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?

Then teach the underlying concepts
Don't copy without citing sources
preview
?

#### Explanation

Explain in detail...

#### Explanation:

I want someone to double check my answer

9
May 20, 2016

The mass increases linearly but radius decreases exponentially, so the result is
$9.8 \frac{m}{s} ^ 2 \cdot 2 \cdot 4 = 78.4 \frac{m}{s} ^ 2$

#### Explanation:

Let's first look at the equation for the force of gravity:

${F}_{g} = G \frac{{m}_{1} {m}_{2}}{r} ^ 2$

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

${F}_{g} = \frac{M}{r} ^ 2$

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 $\frac{1}{2} r$ for $r$. So first start with the full equation:

${F}_{g} = G \frac{{m}_{1} {m}_{2}}{r} ^ 2$

then make the substitutions:

${F}_{g} = G \frac{\left(2 {m}_{1}\right) {m}_{2}}{\frac{1}{2} r} ^ 2$

${F}_{g} = G \frac{\left(2 {m}_{1}\right) {m}_{2}}{\frac{1}{4} r}$

So the numerator increases linearly ($\times 2$ ) but the denominator reduces by an exponential - in this case ($\times 4$ ).

The force of gravity on Earth is roughly $9.8 \frac{m}{s} ^ 2$ but on this other planet, it would be:

$9.8 \frac{m}{s} ^ 2 \cdot 2 \cdot 4 = 78.4 \frac{m}{s} ^ 2$

• An hour ago
• An hour ago
• An hour ago
• An hour ago
• 5 minutes ago
• 6 minutes ago
• 18 minutes ago
• 27 minutes ago
• 46 minutes ago
• An hour ago
• An hour ago
• An hour ago
• An hour ago
• An hour ago