A satellite A of mass m is at a distance of r | vedclass

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

Question

(D) According to Kepler's Third Law of Planetary Motion,the square of the time period $T$ of a satellite is directly proportional to the cube of its o

Vedclass · Accepted Answer

$1:2\sqrt{2}$

Vedclass · Answer

(D) According to Kepler's Third Law of Planetary Motion,the square of the time period $T$ of a satellite is directly proportional to the cube of its orbital radius $r$,i.e.,$T^2 \propto r^3$.This relationship is independent of the mass of the satellite.Given for satellite $A$: $r_A = r$.Given for satellite $B$: $r_B = 2r$.The ratio of their time periods is given by:$\frac{T_A}{T_B} = \left( \frac{r_A}{r_B} \right)^{3/2} = \left( \frac{r}{2r} \right)^{3/2} = \left( \frac{1}{2} \right)^{3/2} = \frac{1}{2^{3/2}} = \frac{1}{2 \cdot 2^{1/2}} = \frac{1}{2\sqrt{2}}$.Thus,the ratio is $1:2\sqrt{2}$.

$A$ satellite $A$ of mass $m$ is at a distance of $r$ from the centre of the earth. Another satellite $B$ of mass $2m$ is at a distance of $2r$ from the earth's centre. Their time periods are in the ratio of:

Similar Questions

Two planets move around the sun. The periodic times and the mean radii of the orbits are $T_1, T_2$ and $r_1, r_2$ respectively. The ratio $T_1/T_2$ is equal to

Suppose there existed a planet that went around the sun twice as fast as the earth. What would be its orbital size as compared to that of the earth?

The period of revolution increases in the order of:

Which of the following orbits is a possible orbit for a planet?

The time period of a geostationary satellite is $24 \; h$,at a height $6 R_{E}$ ($R_{E}$ is the radius of the Earth) from the surface of the Earth. The time period of another satellite whose height is $2.5 R_{E}$ from the surface will be:

Vedclass Test Series

Exam Paper Generator

Online Exam Module