Question

A newly discovered planet has a density eight times the density of the earth and radius twice the radius of the earth. The time taken by 2 kg mass to fall freely through a distance S near the surface of the earth is 1 second. Then the time taken for 4 kg mass to fall freely through the same distance S near the surface of the new planet is :
A]0.25 sec
B]0.5 sec
C]1 sec
D]4 sec

Answer 1

The time taken for mass of 4 kg to fall freely through the same distance S near the surface of the new planet is 0.25 seconds

Therefore, Option (A) is correct.

Explanation:

Let the density of earth is d and radius R

Then the density of the planet will be 8d and radius 2R

Mass of Earth

$M=\frac{4}{3}\pi R^3\times d$

Mass of the planet

$M'=\frac{4}{3}\pi\times (2R)^3\times 8d$

$\implies M'=64M$

On Earth the acceleration due to gravity

$g=\frac{GM}{R^2}$

On the planet, the acceleration due to gravity of the planet

$g'=\frac{GM'}{(2R)^2}$

$\implies g'=\frac{G\times 64M}{4R^2}$

$\implies g'=16g$

On earth a mass of 2 kg travels distance S in 1 second

Therefore, using the second equation of motion

$S=0\times 1+\frac{1}{2}g\times 1^2$

$\implies S=\frac{g}{2}$

$\implies 2S=g$

Now on the planet if the time taken by the mass of 4 kg to fall is t then the acceleration due to the gravity of the planet acting on the body will be g'

Note that the acceleration due to the planets in either case will not depend upon the mass of the objects

Thus,

$S=0\times t+\frac{1}{2}g't^2$

$\implies S=\frac{1}{2}\times 16g\times t^2$

or, $S=8gt^2$

or, $S=8\times 2S\times t^2$

or, $t^2=\frac{1}{16}$

or, $t=\sqrt{\frac{1}{16}}$

or, $t=\frac{1}{4}$ seconds

or, $t=0.25$ seconds

Hope this answer is helpful.

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