Question

Escape velocity from the surface of the earth is V. An object is dropped from infinity to earth, which passes through a smooth tunnel from the surface of the earth to the centre. Find the velocity when it reaches the centre.

Answer 1

Since the object is dropped from infinity on the earth, it has the escape velocity V at the surface of the earth during its motion. How?

Consider a point mass m dropped from a distance x from the center of the earth of radius R and mass M where x >> R. Then the gravitational force of attraction between the point mass and the earth is,

$F=\dfrac {GMm}{x^2}$

Here F is a varying force.

Let the point mass move towards through a small distance 'dx' by this gravitational force of attraction. Then the work done for this is,

$dW=F\cdot (-dx)\\\\\\dW=-\dfrac {GMm}{x^2}\ dx$

The negative sign is because the displacement is opposite to the force of attraction. The displacement is acting towards earth, but the force of attraction is towards the point mass from the earth.

Hence total work done to move the point mass from infinity (x = ∞) to the surface of earth (x = R) will be,

$\displaystyle W=\int\limits_{\infty}^R-\dfrac {GMm}{x^2}\ dx\\\\\\W=-GMm\int\limits_{\infty}^Rx^{-2}\ dx\\\\\\W=-GMm\left [\dfrac {x^{-1}}{-1}\right]_{\infty}^R\\\\\\W=GMm\left (\dfrac {1}{R}-\dfrac {1}{\infty}\right)\\\\\\W=\dfrac {GMm}{R}$

Well, this is the total mechanical energy in the point mass. But since this work is done by the motion of the point mass, we can say that this work done is stored in the point mass as the kinetic energy. Hence,

$\dfrac {1}{2}mu^2=\dfrac {GMm}{R}\\\\\\u=\sqrt {\dfrac {2GM}{R}}$

But, $g=\dfrac {GM}{R^2}\quad\implies\quad GM=gR^2$

Then,

$u=\sqrt {\dfrac {2gR^2}{R}}\\\\\\u=\sqrt {2gR}$

Well, this is the same as escape velocity. $\therefore\ u=V.$

So we can have,

• initial velocity of the object at the surface of the earth = V.

• Since the object is considered to pass from the surface of the earth towards the center, displacement of the object is R.

But the acceleration, g, varies from the surface of the earth with a depth d towards the center as,

$g'=g\left(1-\dfrac {d}{R}\right)$

where g' is the acceleration due to gravity at the center of the earth.

Let the object travel a small depth dd inside the earth by the acceleration and initial velocity. Then, after travelling this small depth dd, by third kinematic equation, the final velocity of the object will be given by,

$v^2-u^2=2g'\cdot dd\\\\\\d(v^2)=2g\left (1-\dfrac {d}{R}\right) dd$

Then the final velocity obtained by the object by traveling from the surface of the earth (d = 0) to the center (d = R) is,

$\displaystyle v^2=2g\int\limits_0^{R}\left (1-\dfrac {d}{R}\right)\ dd\\\\\\v^2=2g\left [\int\limits_0^{R}1\ dd-\dfrac {1}{R}\int\limits_0^{R}d\ dd\right]\\\\\\v^2=2g\left [\left [d\right]_0^{R}-\dfrac {1}{R}\left [\dfrac {d^2}{2}\right]_0^{R}\right]+c\\\\\\v^2=2g\left [\left (R-0\right)-\dfrac {1}{2R}\left (R^2-0^2\right)\right]+c\\\\\\v^2=2g\left [R-\dfrac {R}{2}\right]+c\\\\\\v^2=gR+c$

Here, $c=u^2.$ Then,

$v^2=u^2+gR$

But, $u=V=\sqrt {2gR}.$ Then,

$v^2=2gR+gR\\\\\\v^2=3gR\\\\\\v^2=\dfrac {3}{2}\cdot 2gR\\\\\\v^2=\dfrac {3}{2}V^2\\\\\\v^2=1.5V^2$

Therefore,

$\large\boxed {\mathbf {v=V\sqrt {1.5}}}$

Thus the velocity of the object at the center of the earth is √(1.5) times the escape velocity.