Question

z Assume a planet is a uniform sphere of radiusRthat (somehow) has a narrow radial tunnel through its center. Also assume we can position an apple any where a long the tunnel or outside the sphere. Let${\mathbf{F}}_{\mathbf{R}}$be the magnitude of the gravitational force on the apple when it is located at the planet’s surface. How far from the surface is there a point where the magnitude isrole="math" localid="1657195577959" ${\mathbf{F}}_{\mathbf{R}}$if we move the apple (a) away from the planet and (b) into the tunnel?

Short answer

1. The distance of the apple if it is moved away from the planet when$F_R=\frac{1}{2}{F}_r$ is
   $0.414r$
2. The distance of the apple if it is moved into the tunnel when$F_R=\frac{1}{2}{F}_r$ is,
   $\frac{R}{2}$

Step-by-step solution

The given data

Gravitational force on the apple when it is on surface of planet is$F_r$

Understanding the concept of Newton’s law of gravitation

Using Newton’s law of gravitation, we can write the forces in terms of masses and distances. Using these force equations, we can find the required distances.

Formula:

Gravitational force,$F=\frac{GMm}{r^2}$ (i)

Volume of a sphere, $v=\frac{4}{3}{\mathrm{\pi r}}^3$ (ii)

Density of a material, $p=\frac{M}{v}$ (iii)

a) Calculation of distance on the apple when moved away from planet

Force on the apple at a distance r away from planet can be given by equation (i).

So, when apple is on the surface of planet, force is:

$F_r=\frac{GmM}{r^2}......................\left(a\right)$

When moved away from planet, force$F=F_Y/2$and distance is d, therefore

$\frac{F_r}{2}=\frac{GmM}{d^2}......................\left(b\right)$

From equations (a) & (b), we can write

$$\begin{gathered} \frac{1}{2}=r^2/d^2 \\ d=r\sqrt{2} \end{gathered}$$

This is distance from center of planet

So distance from surface =$(\sqrt{2}-1)r=0.414r$

b) Calculation of the distance of the apple when moved towards tunnel.

Mass of the apple can be written as follows using equations (ii) & (iii),

$M=\frac{4}{3}{\mathrm{\pi r}}^{3}p$

So force on the apple when moved towards tunnel can be given as

$$\begin{gathered} F_r=\frac{4Gmrp}{3} \\ =\frac{4Gmr}{3}\times \frac{M}{4{\mathrm{\pi R}}^3} \\ =\frac{GMmr}{R^3}........................\left(1\right) \end{gathered}$$

But, on surface of planet force is

Equating both equations (1) & (2), we get

Hence, the point from the surface of the planet is at a distance of$\frac{R}{2}$