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?

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

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)

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 =$(\sqrt{2}-1)r=0.414r$

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$\frac{R}{2}$