Physics, asked by shifa1812, 8 months ago

3. 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?
1)4 25 3)2 403​

Answers

Answered by lovebinrp
2

Answer:

4

ExplanationBy applying Keplers third law

Of planetary motion

HOPE THIS HELPS!!!

Attachments:
Answered by sonuvuce
0

The distance of A from  the Sun is 4 times greater than that of B  from the Sun

Option (1) is correct

Explanation:

From Kepler's law we know that if T is the time period and r is the distance of the planet from the Sun then

T^2\propto r^3

Terefore, for the planets A and B

\frac{T_A^2}{T_B^2}=\frac{r_A^3}{r_B^3}

Given that

\frac{T_A}{T_B}=8

\implies 8^2=\frac{r_A^3}{r_B^3}

\implies \frac{r_A^3}{r_B^3}=64

\implies \frac{r_A}{r_B}=\sqrt[3]{64}

\implies \frac{r_A}{r_B}=4

\implies r_A=4r_B

Thus, the distance of A from the Sun is 4 times greater than that of B

Hope this helps.

Know More:

Q: Similar question

Click Here: https://brainly.in/question/7106517

Q: A planet of mass m moves around the sun of mass M in an elliptical orbit .the maximum and minimum distances of the planet from the sun are r1 and r2 respectively .the time period of the planet is proportional to:

Click Here: https://brainly.in/question/2035749

Similar questions