Assume that the earth moves around the sun in | 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 $r$:$T^2

Vedclass · Accepted Answer

$2^{-2 / 3} R$

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 $r$:$T^2 \propto r^3$Since angular speed $\omega = \frac{2\pi}{T}$, we have $T = \frac{2\pi}{\omega}$. Substituting this into the law:$(\frac{2\pi}{\omega})^2 \propto r^3 \Rightarrow \frac{1}{\omega^2} \propto r^3 \Rightarrow r^3 \omega^2 = 	ext{constant}$.For the earth $(E)$ and the planet $(P)$:$r_E^3 \omega_E^2 = r_P^3 \omega_P^2$Given $r_E = R$ and $\omega_P = 2\omega_E$:$R^3 \omega_E^2 = r_P^3 (2\omega_E)^2$$R^3 \omega_E^2 = r_P^3 (4\omega_E^2)$$R^3 = 4 r_P^3$$r_P^3 = \frac{R^3}{4} = \frac{R^3}{2^2}$Taking the cube root on both sides:$r_P = \frac{R}{2^{2/3}} = 2^{-2/3} R$.

Assume that the earth moves around the sun in a circular orbit of radius $R$ and there exists a planet which also moves around the sun in a circular orbit with an angular speed twice as large as that of the earth. The radius of the orbit of the planet is

Similar Questions

State and prove Kepler's second law (Law of Areas) of planetary motion.

The Earth is revolving around the Sun. If the distance of the Earth from the Sun is reduced to $\frac{1}{4}$ of the present distance,then the length of the present day will be reduced by a factor of:

Two planets revolve around the sun with frequencies ${N_1}$ and ${N_2}$ revolutions per year. If their average orbital radii are ${R_1}$ and ${R_2}$ respectively,then ${R_1}/{R_2}$ is equal to:

$A$ geostationary satellite is orbiting the earth at a height of $6R$ above the surface of the earth ($R$ is the radius of the earth). The time period of another satellite at a height of $2.5R$ from the surface of the earth is:

In planetary motion,the areal velocity of the position vector of a planet depends on the angular velocity $(\omega)$ and the distance of the planet from the sun $(r)$. If so,the correct relation for areal velocity is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module