Question

The time period (T) of an artificial satellite of the earth depends on the density (d) of the earth as.
(A) T ? d
(B) T ? sqrt(d)
(C) T ?1/sqrt(d)
(D) T ?1/d

Answer 1

I think it is (C) T ?1/sqrt(d)

Answer 2

The time period of an artificial satellite of the earth depends on the density of the earth as : $T\propto \dfrac{1}{\sqrt{d}}$

Explanation:

The Kepler's law of planetary motion gives the relation between the time period and the semimajor axis. It is given by :

$T^2\propto R^3$

or

$T^2=\dfrac{4\pi ^2R^3}{GM}$......(1)

G is universal gravitational constant

M is mass of Earth

Density,

$d=\dfrac{M}{V}\\\\M=d\times V$, d is density of Earth

Equation (1) becomes :

$T^2=\dfrac{4\pi ^2R^3}{Gd\times V}\\\\\text{since}, V=\dfrac{4}{3}\pi R^3\\\\T^2=\dfrac{4\pi ^2R^3}{Gd\times \dfrac{4}{3}\pi R^3}\\\\T^2=\dfrac{3\pi}{dG}\\\\T=\sqrt{\dfrac{3\pi}{dG}} \\\\T\propto \dfrac{1}{\sqrt{d}}$

It is clear that the time period of an artificial satellite of the earth depends on the density of the earth as : $T\propto \dfrac{1}{\sqrt{d}}$. So, the correct option is (c).

Learn more,

Kepler's law

https://brainly.in/question/6380224