Physics, asked by akshayhk9, 1 year ago

luj 10 uays
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) 4
(b) 5
(c) 6
(d) 8​

Answers

Answered by Anonymous
0

Answer:

8 may be the right answer. not sure buddy

Answered by sonuvuce
0

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

Option (a) 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.

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