I was reading this question 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?


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


2 Answers 2


There are at least two interpretations to this problem:

Per Wikipedia, Jupiter's surface gravity is $2.528$ times Earth's. Thus, if the Earth were $2.528$ times denser, it would have the same surface gravity as Jupiter. The Earth's current density is $5.514$ grams per cubic centimeter, so the new density would be $2.528 \times 5.514$, or about $13.9394$ grams per cubic centimeter. This assumes we change Earth's mass, but not its radius.

@Rob_Jeffries answer assumes the Earth's mass remains constant and the radius changes. If the radius shrinks by a factor of $2$, the volume decreases by a factor of $8$, and the planet becomes 8 times more dense. The surface gravity increases by $4$, since it depends on the radius squared. In general, shrinking the planet's radius by $x$ will increase the density by $x^3$ and the gravity by $x^2$. If we want the gravity $2.528$ times higher, we choose $x = \sqrt{2.528}$ or right around $1.590$. This makes $x^3$ equal to about $4.019$. Multiplying that by Earth's current density of $5.514$, we get $22.16$ grams per cubic centimeter, pretty close to what Rob got.

So, you can't really change the density without changing anything else: either the mass or the volume must change.

  • $\begingroup$ @HDE226868 True, but with a constant radius (and thus constant volume), density and mass are proportional. $\endgroup$
    – user21
    Commented Feb 21, 2015 at 16:06
  • $\begingroup$ I realized my error and deleted my comment just before you posted your reply. My apologies. $\endgroup$
    – HDE 226868
    Commented Feb 21, 2015 at 16:09
  • $\begingroup$ I think it depends how you interpret the question. You can't change density whilst keeping the mass and radius the same! I assumed the mass was what was fixed. The surface gravity depends on $\rho R$. You have changed the mass of the Earth. $\endgroup$
    – ProfRob
    Commented Feb 21, 2015 at 16:10
  • $\begingroup$ @RobJeffries You are correct. I've edited my answer. $\endgroup$
    – user21
    Commented Feb 21, 2015 at 16:26
  • 1
    $\begingroup$ and I +1-ed it, since as a result the OP has the answer they want. $\endgroup$
    – ProfRob
    Commented Feb 21, 2015 at 16:58

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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