A geostationary satellite above the equator i | vedclass

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

Question

(D) According to Kepler's third law,the time period $T$ of a satellite is related to its orbital radius $r$ by $T \propto r^{3/2}$.For the geostationa

Vedclass · Accepted Answer

$(2.33)$

Vedclass · Answer

(D) According to Kepler's third law,the time period $T$ of a satellite is related to its orbital radius $r$ by $T \propto r^{3/2}$.For the geostationary satellite $(GSS)$ with radius $r_1$ and time period $T_1 = 24 	ext{ hours}$,and the second satellite with radius $r_2$ and time period $T_2$,we have:$\frac{T_2}{T_1} = \left(\frac{r_2}{r_1}ight)^{3/2} = \left(\frac{1}{1.21}ight)^{3/2} = \left(\frac{1}{1.1^2}ight)^{3/2} = \frac{1}{1.1^3} = \frac{1}{1.331} \approx \frac{1}{1.33} = \frac{3}{4}$.Thus,$T_2 = \frac{3}{4} T_1 = \frac{3}{4} 	imes 24 = 18 	ext{ hours}$.The angular velocity of the $GSS$ is $\omega_1 = \frac{2\pi}{T_1}$ and the angular velocity of the second satellite is $\omega_2 = \frac{2\pi}{T_2}$.Since the second satellite moves in the opposite direction,their relative angular velocity is $\omega_{rel} = \omega_1 + \omega_2$.The time period $t_0$ as measured from the $GSS$ is $t_0 = \frac{2\pi}{\omega_1 + \omega_2} = \frac{2\pi}{\frac{2\pi}{T_1} + \frac{2\pi}{T_2}} = \frac{T_1 T_2}{T_1 + T_2}$.Substituting the values: $t_0 = \frac{24 	imes 18}{24 + 18} = \frac{432}{42} = \frac{72}{7} 	ext{ hours}$.Given $t_0 = \frac{24}{p}$,we have $\frac{24}{p} = \frac{72}{7}$,which gives $p = \frac{24 	imes 7}{72} = \frac{7}{3} \approx 2.33$.

Similar Questions

The time period of polar satellites is about ..........

$A$ satellite is orbiting just above the surface of a planet of density $\rho$ with a periodic time $T$. The quantity $T^2 \rho$ is equal to ($G=$ universal gravitational constant).

An object is projected horizontally with speed $\frac{1}{2} \sqrt{\frac{GM}{R}}$ from a point at height $3R$ (where $R$ is the radius and $M$ is the mass of the Earth). The object will:

If $\rho$ is the density of the planet,the time period of a nearby satellite is given by

$A$ satellite $S$ is moving in an elliptical orbit around the Earth. The mass of the satellite is very small compared to the mass of the Earth.

Vedclass Test Series

Exam Paper Generator

Online Exam Module