If a satellite orbiting the Earth is 9 times | vedclass

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

Question

(A) According to Kepler's Third Law of Planetary Motion, the square of the time period $(T)$ is proportional to the cube of the orbital radius $(R)$:

Vedclass · Accepted Answer

$1 	ext{ day}$

Vedclass · Answer

(A) According to Kepler's Third Law of Planetary Motion, the square of the time period $(T)$ is proportional to the cube of the orbital radius $(R)$: $T^2 \propto R^3$.Let $T_m$ and $R_m$ be the time period and orbital radius of the Moon, and $T_s$ and $R_s$ be the time period and orbital radius of the satellite.Given: $R_s = R_m / 9$ and $T_m = 27 	ext{ days}$.Using the ratio formula: $\left(\frac{T_m}{T_s}ight)^2 = \left(\frac{R_m}{R_s}ight)^3$.Substituting the values: $\left(\frac{27}{T_s}ight)^2 = \left(\frac{R_m}{R_m / 9}ight)^3 = (9)^3 = 729$.Taking the square root on both sides: $\frac{27}{T_s} = \sqrt{729} = 27$.Therefore, $T_s = \frac{27}{27} = 1 	ext{ day}$.

If a satellite orbiting the Earth is $9$ times closer to the Earth than the Moon, what is the time period of rotation of the satellite? Given rotational time period of Moon $= 27 \text{ days}$ and gravitational attraction between the satellite and the moon is neglected.

Similar Questions

$A$ planet of mass $m$ moves around the sun of mass $M$ in an elliptical orbit. The maximum and minimum distances of the planet from the sun are $r_1$ and $r_2$ respectively. The time period of the planet is proportional to ..........

$A$ planet moving around the Sun sweeps out an area $A_{1}$ in $2$ days,$A_{2}$ in $3$ days,and $A_{3}$ in $6$ days. Then,the relation between $A_{1}, A_{2}$,and $A_{3}$ is

The time period of a satellite of Earth is $24 \text{ hours}$. If the separation between the Earth and the satellite is decreased to one-fourth of the previous value, then its new time period will become: (in $\text{ hours}$)

If a body describes a circular motion under an inverse square field,the time taken to complete one revolution $T$ is related to the radius of the circular orbit $r$ as:

When a satellite moves around the earth in a certain orbit,the quantity which remains constant is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module