A satellite in a circular orbit of radius R has a period of 4 h. Another satellite with orbital radius 3 R around the same planet will have a period (in hours)

Question

A satellite in a circular orbit of radius R has a period of 4 h. Another satellite with orbital radius 3 R around the same planet will have a period (in hours) (A) 16 (B) 4 (C) 4 √27 (D) 4 √8

Tardigrade Admin · Accepted Answer

According to Kepler's third law T2 propto R3 ⇒ (T2/T1)=((R2/R1))3 / 2 ∴ (T2/T1)=((3 R/R))3 / 2 ⇒ (T2/T1)=√27 ⇒ T2=√27 T1=√27 × 4 =4 √27 h

Q. A satellite in a circular orbit of radius $R$ has a period of $4 h$ . Another satellite with orbital radius $3 R$ around the same planet will have a period (in hours)

16

4

$427$

$48$

According to Kepler's third law
$T^{2} \propto R^{3}$
$\Rightarrow \frac{T _{2}}{T _{1}} = (\frac{R _{2}}{R _{1}})^{3/2}$
$∴ \frac{T _{2}}{T _{1}} = (\frac{3 R}{R})^{3/2}$
$\Rightarrow \frac{T _{2}}{T _{1}} = 27$
$\Rightarrow T_{2} = 27 T_{1} = 27 \times 4$
$= 4 27 h$

Q. A satellite in a circular orbit of radius R has a period of 4h. Another satellite with orbital radius 3R around the same planet will have a period (in hours)

16

4

427​

48​

According to Kepler's third law T2∝R3 ⇒T1​T2​​=(R1​R2​​)3/2 ∴T1​T2​​=(R3R​)3/2 ⇒T1​T2​​=27​ ⇒T2​=27T1​​=27​×4 =427h​

Q. A satellite in a circular orbit of radius $R$ has a period of $4 h$ . Another satellite with orbital radius $3 R$ around the same planet will have a period (in hours)

$427$

$48$

According to Kepler's third law
$T^{2} \propto R^{3}$
$\Rightarrow \frac{T _{2}}{T _{1}} = (\frac{R _{2}}{R _{1}})^{3/2}$
$∴ \frac{T _{2}}{T _{1}} = (\frac{3 R}{R})^{3/2}$
$\Rightarrow \frac{T _{2}}{T _{1}} = 27$
$\Rightarrow T_{2} = 27 T_{1} = 27 \times 4$
$= 4 27 h$