A planet has an orbital radius twice that of | 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 directly proportional to the cube of the semi-major axi

Vedclass · Accepted Answer

$2.8$

Vedclass · Answer

(B) 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 $(R)$ of the orbit: $T^2 \propto R^3$.Let $T_1$ and $R_1$ be the time period and orbital radius of the Earth,and $T_2$ and $R_2$ be the time period and orbital radius of the planet.Given: $T_1 = 1 	ext{ year}$,$R_1 = R$,and $R_2 = 2R$.Using the ratio formula: $\frac{T_2}{T_1} = \left( \frac{R_2}{R_1} ight)^{3/2}$.Substituting the values: $T_2 = 1 	imes \left( \frac{2R}{R} ight)^{3/2} = 2^{3/2}$.Calculating the value: $2^{3/2} = \sqrt{2^3} = \sqrt{8} \approx 2.828 	ext{ years}$.Rounding to the nearest option,the time period is $2.8 	ext{ years}$.

$A$ planet has an orbital radius twice that of the Earth's orbital radius. The time period of the planet is .......... $years$.

Similar Questions

$A$ satellite is launched into a circular orbit of radius $R$ around the Earth. $A$ second satellite is launched into an orbit of radius $1.02 R$. The period of the second satellite is larger than the first one by approximately ........ $\%$

If a new planet is discovered rotating around the Sun with an orbital radius double that of Earth,what will be its time period (in Earth's days)?

The period of revolution of planet $A$ around the sun is $8$ times that of $B$. The distance of $A$ from the sun is how many times greater than that of $B$ from the sun?

$A$ satellite $A$ of mass $m$ is at a distance of $r$ from the centre of the earth. Another satellite $B$ of mass $2m$ is at a distance of $2r$ from the earth's centre. Their time periods are in the ratio of:

The mean radius of the earth's orbit around the sun is $1.5 \times 10^{11} \ m$. The mean radius of the orbit of mercury around the sun is $6 \times 10^{10} \ m$. The mercury will rotate around the sun in

Vedclass Test Series

Exam Paper Generator

Online Exam Module