<p>To simplify things, I will assume that we have a spherical body and that we are considering Newtonian Physics.</p>

<p>If we have two bodies of the same mass but different radius, as long as we are outside of both bodies, and at the same distance from their centres, the force of gravity will be the same:
<span class="math-container">$$
F = G \frac{m M_i}{r_i^2}
$$</span>
where <span class="math-container">$m$</span> is "our" mass, <span class="math-container">$M_i$</span> is the mass of body <span class="math-container">$i$</span> and <span class="math-container">$r_i$</span> is the distance between our centre and the centre of the body, and <span class="math-container">$6.67408 × 10^{-11} m^3 kg^{-1} s^{-2}$</span> is the Gravitational Constant.</p>

<p>So, there is no dependence on the density of the body, only on its total mass.</p>

<hr>

<p>However, if you are thinking about the gravitational acceleration <span class="math-container">$g$</span> at the surface of different planets, not only the masses may differ, but also the distance to their centre. Note that <span class="math-container">$g$</span> is given by:
<span class="math-container">$$
g = G \frac{M_i}{r_i^2}
$$</span></p>

<p>So, computing for Earth we have that <span class="math-container">$M_{Earth} = 5.972 \times 10^{24} kg$</span> and <span class="math-container">$r_{Earth} = 6.371 \times 10^6 m$</span>, which gives, <span class="math-container">$g_{Earth} \approx 9.820 m s^{-2}$</span>, slightly different from the <a href="https://en.wikipedia.org/wiki/Standard_gravity" rel="noreferrer">standard gravity</a>: <span class="math-container">$9.80665 m s^{-2}$</span>.</p>

<p>The same computation for Mars, which has <span class="math-container">$M_{Mars} = 6.39 \times 10^{23} kg$</span> and <span class="math-container">$r_{Mars} = 3.3895 \times 10^6 m$</span>, gives <span class="math-container">$g_{Mars} \approx 3.71 m s^{-2}$</span>. So, since Mars has, roughly, <span class="math-container">$1/10$</span> of Earth's mass and <span class="math-container">$1/2$</span> of the radius, the acceleration of it's gravity is around <span class="math-container">$40\%$</span> of Earth's acceleration.</p>

CC BY-SA 4.0
2020-01-01T17:28:02.433Z
2020-01-01T17:28:02.433Z
Ertxiem - reinstate Monica
<p>the more concentrated the mass, the more powerful is gravity? or does it depend exclusively on the total mass? example: 1 <span class="math-container">$\small\mathsf{cm^3}$</span> material A = 1000 kg, 20 <span class="math-container">$\small\mathsf{cm^3}$</span> material B = 2000 kg
which of the two generates more powerful gravitational force?</p>

<p>I apologize for my English, I am not a native speaker, and I used google translate</p>

<astrophysics>
the more concentrated the mass, the more powerful is gravity?
Anton Ivanov
# the more concentrated the mass, the more powerful is gravity?

(Asked by: username_0 on 2020-01-01)

the more concentrated the mass, the more powerful is gravity? or does it depend exclusively on the total mass? example: 1 $\small\mathsf{cm^3}$ material A = 1000 kg, 20 $\small\mathsf{cm^3}$ material B = 2000 kg
which of the two generates more powerful gravitational force?

I apologize for my English, I am not a native speaker, and I used google translate

**Tags:** astrophysics

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## 1 answers

### Answer by username_1 on 2020-01-01. Score: 5 | Accepted answer

To simplify things, I will assume that we have a spherical body and that we are considering Newtonian Physics.

If we have two bodies of the same mass but different radius, as long as we are outside of both bodies, and at the same distance from their centres, the force of gravity will be the same:
$$
F = G \frac{m M\_i}{r\_i^2}
$$
where $m$ is "our" mass, $M\_i$ is the mass of body $i$ and $r\_i$ is the distance between our centre and the centre of the body, and $6.67408 × 10^{-11} m^3 kg^{-1} s^{-2}$ is the Gravitational Constant.

So, there is no dependence on the density of the body, only on its total mass.

---

However, if you are thinking about the gravitational acceleration $g$ at the surface of different planets, not only the masses may differ, but also the distance to their centre. Note that $g$ is given by:
$$
g = G \frac{M\_i}{r\_i^2}
$$

So, computing for Earth we have that $M\_{Earth} = 5.972 \times 10^{24} kg$ and $r\_{Earth} = 6.371 \times 10^6 m$, which gives, $g\_{Earth} \approx 9.820 m s^{-2}$, slightly different from the [standard gravity](https://en.wikipedia.org/wiki/Standard_gravity): $9.80665 m s^{-2}$.

The same computation for Mars, which has $M\_{Mars} = 6.39 \times 10^{23} kg$ and $r\_{Mars} = 3.3895 \times 10^6 m$, gives $g\_{Mars} \approx 3.71 m s^{-2}$. So, since Mars has, roughly, $1/10$ of Earth's mass and $1/2$ of the radius, the acceleration of it's gravity is around $40\%$ of Earth's acceleration.


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