The time period of a satellite of Earth is 5, | vedclass

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

Question

(C) According to Kepler's third law of planetary motion,the square of the time period $(T)$ is proportional to the cube of the semi-major axis $(r)$,i

Vedclass · Accepted Answer

$40$

Vedclass · Answer

(C) According to Kepler's third law of planetary motion,the square of the time period $(T)$ is proportional to the cube of the semi-major axis $(r)$,i.e.,$T^2 \propto r^3$ or $T \propto r^{3/2}$.Given the initial time period $T_1 = 5\, h$ and the initial separation $r_1 = r$.The new separation is $r_2 = 4r$.Using the ratio formula: $\frac{T_2}{T_1} = \left(\frac{r_2}{r_1}ight)^{3/2}$.Substituting the values: $\frac{T_2}{5} = \left(\frac{4r}{r}ight)^{3/2}$.$\frac{T_2}{5} = (4)^{3/2} = (2^2)^{3/2} = 2^3 = 8$.Therefore,$T_2 = 5 	imes 8 = 40\, h$.

The time period of a satellite of Earth is $5\, h$. If the separation between the centre of the Earth and the satellite is increased to $4$ times the previous value,the new time period will become ....... $h$.

Similar Questions

$A$ 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:

$A$ satellite moves around the Earth in a circular orbit of radius $R$ making one revolution per day. $A$ second satellite moving in a circular orbit moves around the Earth once in $8$ days. The radius of the orbit of the second satellite is

Describe the model and theory of planetary motions presented in the earliest of times.

The torque on a planet about the centre of the Sun is ..............

The period of a planet around the sun is $8$ times that of earth. The ratio of the radius of the planet's orbit to the radius of the earth's orbit is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module