Two satellites A and B having ratio of masses | vedclass

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

Question

(D) The total energy $E$ of a satellite of mass $m$ revolving in an orbit of radius $r$ around a planet of mass $M$ is given by $E = -\frac{GMm}{2r}$.

Vedclass · Accepted Answer

$12: 1$

Vedclass · Answer

(D) The total energy $E$ of a satellite of mass $m$ revolving in an orbit of radius $r$ around a planet of mass $M$ is given by $E = -\frac{GMm}{2r}$.From this expression,we can see that $E \propto \frac{m}{r}$.Given the mass ratio $\frac{m_A}{m_B} = \frac{3}{1}$ and the radius ratio $\frac{r_A}{r_B} = \frac{r}{4r} = \frac{1}{4}$.Therefore,the ratio of the total energies is $\frac{E_A}{E_B} = \frac{m_A}{m_B} 	imes \frac{r_B}{r_A}$.Substituting the given values: $\frac{E_A}{E_B} = \frac{3}{1} 	imes \frac{4r}{r} = \frac{12}{1}$.Thus,the ratio is $12: 1$.

Two satellites $A$ and $B$ having ratio of masses $3: 1$ are revolving in circular orbits of radii $r$ and $4r$. The ratio of total energy of satellites $A$ to that of $B$ is

Similar Questions

$A$ satellite moves in a stable circular orbit around the Earth if (where $V_{H}$,$V_{c}$,and $V_{e}$ are the horizontal velocity,critical velocity,and escape velocity respectively):

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

The orbital period of a geostationary satellite is (in $\,h$)

$A$ particle moves in a circular orbit of radius $r$ under a central attractive force $F = -\frac{k}{r}$,where $k$ is a constant. The periodic time of this motion is proportional to:

Consider a satellite going round the Earth in an orbit. Which of the following statements is wrong?

Vedclass Test Series

Exam Paper Generator

Online Exam Module