Given below are two statements:Statement I: A | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The time period of a satellite orbiting close to the Earth's surface is given by $T = 2\pi\sqrt{\frac{R_e^3}{GM}}$.Since the mass of the Earth $M

Vedclass · Accepted Answer

Both Statement $I$ and Statement $II$ are true

Vedclass · Answer

(B) The time period of a satellite orbiting close to the Earth's surface is given by $T = 2\pi\sqrt{\frac{R_e^3}{GM}}$.Since the mass of the Earth $M = \rho \cdot \frac{4}{3}\pi R_e^3$, where $\rho$ is the density of the Earth, we can substitute $M$ into the formula:$T = 2\pi\sqrt{\frac{R_e^3}{G(\rho \cdot \frac{4}{3}\pi R_e^3)}} = 2\pi\sqrt{\frac{3}{4\pi G\rho}}$.This shows that $T$ depends on the density $\rho$ of the Earth. Thus, Statement $I$ is true.For a satellite very close to the surface, the gravitational force provides the centripetal force, leading to $g = \frac{GM}{R_e^2}$.Substituting $GM = gR_e^2$ into the general time period formula $T = 2\pi\sqrt{\frac{R_e^3}{GM}}$, we get $T = 2\pi\sqrt{\frac{R_e^3}{gR_e^2}} = 2\pi\sqrt{\frac{R_e}{g}}$.Thus, Statement $II$ is true.

Similar Questions

Imagine a light planet revolving around a very massive star in a circular orbit of radius $R$ with a period of revolution $T$. If the gravitational force of attraction between the planet and the star is proportional to $R^{-5/2}$,then ${T^2}$ is proportional to:

$A$ geostationary satellite is orbiting the earth at a height $5R$ above the surface of the earth,where $R$ is the radius of the earth. What is the time period (in hours) of another satellite orbiting at a height of $2R$ from the surface of the earth?

Two satellites of masses $m$ and $3\,m$ revolve around the earth in circular orbits of radii $r$ and $3r$ respectively. The ratio of orbital speeds of the satellites respectively is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module