Two planets are at mean distances d_1 and d_2 | vedclass

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

Question

(B) According to Kepler's Third Law of Planetary Motion,the square of the time period $T$ is proportional to the cube of the mean distance $d$ from th

Vedclass · Accepted Answer

$n_1^2 d_1^3 = n_2^2 d_2^3$

Vedclass · Answer

(B) According to Kepler's Third Law of Planetary Motion,the square of the time period $T$ is proportional to the cube of the mean distance $d$ from the sun: $T^2 \propto d^3$.Since frequency $n$ is the reciprocal of the time period $(n = 1/T)$,we have $T = 1/n$.Substituting this into the law: $(1/n)^2 \propto d^3$,which implies $1/n^2 \propto d^3$.Rearranging this gives $n^2 d^3 = 	ext{constant}$.Therefore,for two planets,we have $n_1^2 d_1^3 = n_2^2 d_2^3$.

Two planets are at mean distances $d_1$ and $d_2$ from the sun,and their orbital frequencies are $n_1$ and $n_2$ respectively. Then:

Similar Questions

The period of a satellite in a circular orbit of radius $R$ is $T$. The period of another satellite in a circular orbit of radius $4R$ 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 $r$. The correct relation for areal velocity is:

$A$ planet revolves around the sun in an elliptical orbit. The linear speed of the planet will be maximum at

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:

Vedclass Test Series

Exam Paper Generator

Online Exam Module