A planet takes 200 days to complete one revol | 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 directly proportional to the cube of the semi-major axi

Vedclass · Accepted Answer

$25$

Vedclass · Answer

(A) According to Kepler's third law of planetary motion,the square of the time period $(T)$ is directly proportional to the cube of the semi-major axis $(r)$: $T^2 \propto r^3$.Let the initial time period be $T_1 = 200 	ext{ days}$ and the initial distance be $r_1 = r$.Let the new time period be $T_2$ and the new distance be $r_2 = \frac{r}{4}$.Using the relation $\frac{T_1^2}{r_1^3} = \frac{T_2^2}{r_2^3}$,we get:$\frac{(200)^2}{r^3} = \frac{T_2^2}{(\frac{r}{4})^3}$$T_2^2 = (200)^2 	imes \frac{(\frac{r}{4})^3}{r^3}$$T_2^2 = (200)^2 	imes \frac{r^3}{64 	imes r^3}$$T_2^2 = \frac{(200)^2}{64}$Taking the square root of both sides:$T_2 = \frac{200}{\sqrt{64}}$$T_2 = \frac{200}{8}$$T_2 = 25 	ext{ days}$.

$A$ planet takes $200$ days to complete one revolution around the Sun. If the distance of the planet from the Sun is reduced to one-fourth of the original distance,how many days will it take to complete one revolution?

Similar Questions

The period of revolution increases in the order of:

The motion of planets in the solar system is an example of the conservation of:

The radius of the Martian orbit around the Sun is about $4$ times the radius of the orbit of Mercury. The Martian year is $687$ Earth days. What is the length of $1$ year on Mercury?

The motion of satellites in the solar system is an example of which of the following?

Vedclass Test Series

Exam Paper Generator

Online Exam Module