The orbit of a geostationary satellite is cir | vedclass

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

Question

(D) The gravitational force provides the necessary centripetal force for the circular orbit of the satellite: $\frac{GMm}{(R+h)^{2}} = m \omega^{2}(R+

Vedclass · Accepted Answer

$(ii), (iii)$ and $(iv)$

Vedclass · Answer

(D) The gravitational force provides the necessary centripetal force for the circular orbit of the satellite: $\frac{GMm}{(R+h)^{2}} = m \omega^{2}(R+h)$.Here,$G$ is the gravitational constant,$M$ is the mass of the Earth,$m$ is the mass of the satellite,$R$ is the radius of the Earth,and $h$ is the height of the satellite above the surface.Simplifying for angular velocity $\omega$: $\omega = \sqrt{\frac{GM}{(R+h)^{3}}}$.Since $\omega = \frac{2\pi}{T}$,the time period $T$ is given by $T = 2\pi \sqrt{\frac{(R+h)^{3}}{GM}}$.The radius of the orbit is $r = R+h$.Thus,$T = 2\pi \sqrt{\frac{r^{3}}{GM}}$.From this expression,$T$ depends on the mass of the Earth $(M)$,the radius of the orbit $(r)$,and the height $(h)$ (since $r = R+h$). It does not depend on the mass of the satellite $(m)$.Therefore,the correct factors are $(ii), (iii),$ and $(iv)$.

The orbit of a geostationary satellite is circular. The time period of the satellite depends on $(i)$ mass of the satellite,$(ii)$ mass of the earth,$(iii)$ radius of the orbit,and $(iv)$ height of the satellite from the surface of the earth.

Similar Questions

$A$ planet is revolving around the sun in a circular orbit with a radius $r$. The time period is $T$. If the force between the planet and the star is proportional to $r^{-3/2}$,then the square of the time period is proportional to

What is the orbital velocity of a satellite? Derive its equation.

Three identical bodies of equal mass $M$ each are moving along a circle of radius $R$ under the action of their mutual gravitational attraction. The speed of each body is

What is the period of revolution of an Earth satellite? Ignore the height of the satellite above the surface of the Earth.
Given:
$(1)$ The value of gravitational acceleration $g = 10 \ ms^{-2}$.
$(2)$ Radius of Earth $R_E = 6400 \ km$. Take $\pi = 3.14$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module