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?

Question

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) 2 (D) 3

Tardigrade Admin · Accepted Answer

Period of revolution of planet A(TA)=8 TB. According to Kepler's III law of planetary motion T2 propto R3. Therefore ((rA/rB))3=((TA/TB))2=((8 TB/TB))2=64 or (rA/rB)=4 or rA=4 rB.

Q. 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?

4

5

2

3

Period of revolution of planet $A (T_{A}) = 8 T_{B}$ .
According to Kepler's III law of planetary motion $T^{2} \propto R^{3}$ .
Therefore $(\frac{r _{A}}{r _{B}})^{3} = (\frac{T _{A}}{T _{B}})^{2} = (\frac{8 T _{B}}{T _{B}})^{2} = 64$
or $\frac{r _{A}}{r _{B}} = 4$
or $r_{A} = 4 r_{B}$ .

Q. 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?

4

5

2

3

Period of revolution of planet A(TA​)=8TB​. According to Kepler's III law of planetary motion T2∝R3. Therefore (rB​rA​​)3=(TB​TA​​)2=(TB​8TB​​)2=64 or rB​rA​​=4 or rA​=4rB​.

Q. 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?

Period of revolution of planet $A (T_{A}) = 8 T_{B}$ .
According to Kepler's III law of planetary motion $T^{2} \propto R^{3}$ .
Therefore $(\frac{r _{A}}{r _{B}})^{3} = (\frac{T _{A}}{T _{B}})^{2} = (\frac{8 T _{B}}{T _{B}})^{2} = 64$
or $\frac{r _{A}}{r _{B}} = 4$
or $r_{A} = 4 r_{B}$ .