Question

Show that on a hypothetical planet having half the diameter of the Earth but twice its density, the acceleration of free fall is the same as on Earth.

Answer 1

Answer:

Steps include:

• Find the ratio between the mass of this hypothetical planet and the mass of planet earth.
• Calculate the ratio between the acceleration of free fall (gravitational field strength) near the surface of that hypothetical planet and near the surface of the earth.

Assumption: both planet earth and this hypothetical planet are spheres of uniform density.

Explanation:

Mass of that hypothetical planet

Let $M_0$, $r_0$, and $\rho_0$ denote the mass, radius, and density of planet earth.

The radius and density of this hypothetical planet would be $(1/2)\, r_0$ and $2\, \rho_0$, respectively.

The volume of a sphere is $\displaystyle V = \frac{4}{3}\, \pi\, r^3$.

Therefore:

• The volume of planet earth would be $\displaystyle V_0 = \frac{4}{3}\, \pi\, {r_0}^{3}$.

• The volume of this hypothetical planet would be: $\displaystyle V = \frac{4}{3}\, \pi\, \left(\frac{r_0}{2}\right)^3 = \frac{1}{6}\, \pi\, {r_0}^3$.

The mass of an object is the product of its volume and density. Hence, the mass of the earth could also be represented as:

$\displaystyle M_0 = \rho_0\, V_0 = \frac{4}{3}\, \pi\, {r_0}^{3}\, \rho_0$.

In comparison, the mass of this hypothetical planet would be:

$\begin{aligned}M &= \rho\, V\\ &= (2\, \rho_0) \cdot \left(\frac{1}{6}\, \pi\, {r_0}^3\right) \\ &= \frac{1}{3}\, \pi\, {r_0}^{3}\, \rho_0 \end{aligned}$.

Compare these two expressions. Notice that $\displaystyle M = \frac{1}{4}\, M_0$. In other words, the mass of the hypothetical planet is one-fourth the mass of planet earth.

Gravitational field strength near the surface of that hypothetical planet

The acceleration of free fall near the surface of a planet is equal to the gravitational field strength at that very position.

Consider a sphere of uniform density. Let the mass and radius of that sphere be $M$ and $r$, respectively. Let $G$ denote the constant of universal gravitation. Right next to the surface of that sphere, the strength of the gravitational field of that sphere would be:

$\displaystyle g = \frac{G \cdot M}{r^2}$.

(That's the same as if all the mass of that sphere were concentrated at a point at the center of that sphere.)

Assume that both planet earth and this hypothetical planet are spheres of uniform density.

Using this equation, the gravitational field strength near the surface the earth would be:

$\displaystyle g_0 &= \frac{G \cdot M_0}{{r_0}^2}$.

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On the other hand, the gravitational field strength near the surface of that hypothetical planet would be:

$\begin{aligned}g &= \frac{G \cdot (M_0/4)}{{(r_0/2)}^2} = \frac{G \cdot M_0}{{r_0}^2}\end{aligned}$.

Notice that these two expressions are equal. Therefore, the gravitational field strength (and hence the acceleration of free fall) would be the same near the surface of the earth and near the surface of that hypothetical planet.

As a side note, both $g$ and $g_0$ could be expressed in terms of $\rho_0$ and $r_0$ alongside the constants $\pi$ and $G$:

$\begin{aligned}g &= \frac{G \cdot M}{r^2}\\ &= \frac{G \cdot (1/3)\, \pi\, {r_0}^3 \, \rho_0}{(r_0/2)^2} = \frac{4}{3}\,\pi\, G\, r_0\, \rho_0\end{aligned}$.

$\begin{aligned}g_0 &= \frac{G \cdot M_0}{{r_0}^2}\\ &= \frac{G \cdot (4/3)\, \pi\, {r_0}^3 \, \rho_0}{{r_0}^2} = \frac{4}{3}\,\pi\, G\, r_0\, \rho_0\end{aligned}$.