Question

In the rough approximation that the density of the Earth is uniform throughout its interior, the gravitational field strength (force per unit mass) inside the Earth at a distance rfrom the centre is$\frac{gr}{R}$ , whereis the radius of the Earth. (In actual fact, the outer layers of rock have lower density than the inner core of molten iron.) Using the uniform-density approximation, calculate the amount of energy required to move a massfrom the centre of the Earth to the surface. Compare with the amount of energy required to move the mass from the surface of the Earth to a great distance away.

Short answer

The energy required is $\frac{1}{2}mgR$and equals the half energy required to move the mass from the surface of the Earth to a great distance away

Step-by-step solution

Given Data

The gravitational force per unit mass placed at a point in a gravitational field is defined as the gravitational field strength at that point.

The strength of the gravitational field is defined as

$\stackrel{-}{g}=\frac{gr}{R}$.............(1)

Here,

g = Acceleration due to gravity

r = The distance between the earth's centre and the assumed point of view.

R = Radius of the earth

Concept of the gravitational force and work

Consider a particle with mass mand a distance rfromthe earth's centre.

The particle is subjected to the gravitational force

$F=mg=\frac{mgr}{R}$

The amount of effort required to move a unit mass via an endlessly small displacement dris called work.

Determine the work done

The amount of energy required to move a mass (m) from the center of the earth to the surface. To move the mass, we should apply a work done on the mass where this work done is given by

$W={\int }_{}^{}\overrightarrow{F}dr$

Where F is the force applied by the earth on the mass and it is given by

$\overrightarrow{F}=m\overrightarrow{g}=m\frac{gr}{R}$

Now let us plug the expression $\overrightarrow{F}$into equation (1) and integrate over the radius of the earth $\left(0\to R\right)$

$$\begin{gathered} W={\int }_0^R\frac{mg}{R}r.dr \\ W=\frac{mg}{R}{\int }_0^{R}r.dr \\ W=\frac{mg}{R}{\left[\frac{r^2}{2}\right]}_0^R \\ W=\frac{1}{2}mgR \end{gathered}$$

In the second part, use equation (1) and integrate over infinity$\left(R\to \infty \right)$to get the energy required to move the mass to a great distance away

$$\begin{gathered} W=\frac{mg}{R}{\int }_0^{R}r.dr \\ W=\frac{mg}{R}{\left[\frac{r^2}{2}\right]}_R^{\infty } \\ W=mgR \end{gathered}$$

The energy required to move a mass from the centre of the Earth to its surface is half the amount of energy required to move the mass from the surface of the Earth to a great distance away.

Therefore, the energy required is $\frac{1}{2}mgR$and equals the half energy required to move the mass from the surface of the Earth to a great distance away.