<p>You only need two formulae. Gravitational field of a spherically symmetric mass distribution is given by 
$$g = \frac{GM}{R^2},$$
where $M$ is the mass inside a radius $R$. The second formula is the average density of a sphere is its mass divided by its volume, hence
$$\rho = \frac{M}{(4/3)\pi R^3}$$</p>

<p>These two formulae can obviously be put together to give the gravitational field as a function of mass and density. 
$$ g = \frac{GM}{(3M/4\pi \rho)^{2/3}}$$
$$ \rho = \frac{3}{4\pi} g^{3/2} M^{-1/2} G^{-3/2}$$</p>

<p>Using $g=24.8$ m/s$^{2}$ for the surface gravity of Jupiter and $M=6\times10^{24}$ kg for the (unchanged) mass of the Earth. We obtain $\rho = 22096$ kg/m$^{3}$.</p>

<p>Note that my answer assumes that the <em>mass</em> of the Earth is fixed. If you instead change the mass and leave the radius fixed:
$$ \rho = \frac{3g}{4\pi G R}$$
which gives $\rho = 13930$ kg/m$^{3}$.</p>

<p>You can't leave mass and radius fixed!</p>

CC BY-SA 3.0
2015-02-21T12:29:00.540Z
2015-02-21T16:22:18.097Z
2015-02-21T16:22:18.097Z
ProfRob
<p>I was reading this <a href="https://astronomy.stackexchange.com/questions/2315/how-small-can-a-planet-be-and-still-have-earth-like-gravity?rq=1">question</a> about <code>how small could a planet be while having earth-like gravitational pull</code>. </p>

<p>This got me thinking, how dense would planet Earth have to be to have the same gravitational pull as Jupiter while all the other factors staying the same (even if it is impossible in the real world)?</p>

<p>If there are any formulas could you please explain them so I understand them please?</p>

<p><strong>Edit</strong></p>

<p>Sorry about some of the confusion but I meant keeping everything <strong>but</strong> the mass the same.</p>

<gravity><earth><jupiter>
How dense would planet earth have to be to have the same gravity as Jupiter?
iProgram
# How dense would planet earth have to be to have the same gravity as Jupiter?

(Asked by: username_0 on 2015-02-21)

I was reading this [question](https://astronomy.stackexchange.com/questions/2315/how-small-can-a-planet-be-and-still-have-earth-like-gravity?rq=1) about `how small could a planet be while having earth-like gravitational pull`.

This got me thinking, how dense would planet Earth have to be to have the same gravitational pull as Jupiter while all the other factors staying the same (even if it is impossible in the real world)?

If there are any formulas could you please explain them so I understand them please?

**Edit**

Sorry about some of the confusion but I meant keeping everything **but** the mass the same.

**Tags:** gravity, earth, jupiter


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## 1 answers

### Answer by username_1 on 2015-02-21. Score: 4

You only need two formulae. Gravitational field of a spherically symmetric mass distribution is given by
$$g = \frac{GM}{R^2},$$
where $M$ is the mass inside a radius $R$. The second formula is the average density of a sphere is its mass divided by its volume, hence
$$\rho = \frac{M}{(4/3)\pi R^3}$$

These two formulae can obviously be put together to give the gravitational field as a function of mass and density.
$$ g = \frac{GM}{(3M/4\pi \rho)^{2/3}}$$
$$ \rho = \frac{3}{4\pi} g^{3/2} M^{-1/2} G^{-3/2}$$

Using $g=24.8$ m/s$^{2}$ for the surface gravity of Jupiter and $M=6\times10^{24}$ kg for the (unchanged) mass of the Earth. We obtain $\rho = 22096$ kg/m$^{3}$.

Note that my answer assumes that the *mass* of the Earth is fixed. If you instead change the mass and leave the radius fixed:
$$ \rho = \frac{3g}{4\pi G R}$$
which gives $\rho = 13930$ kg/m$^{3}$.

You can't leave mass and radius fixed!


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