Satellite revolve in radius R has a time period of revolution T. Find time period of a satellite if orbital radius is 9R ?

Question

Satellite revolve in radius R has a time period of revolution T. Find time period of a satellite if orbital radius is 9R ? (A) 3√3T (B) 9T (C) 27 T (D) 9√3T

Tardigrade Admin · Accepted Answer

According to Kepler's law of periods T2 propto R3 So, ⇒ (T1/T2)=((R1/R2))(3/2) ⇒ (T/T2)=((R/9R))(3/2) =(1/27) [∵ T1=T] T2=27T

Q. Satellite revolve in radius $R$ has a time period of revolution $T$ . Find time period of a satellite if orbital radius is $9 R$ ?

$33 T$

9T

27 T

$93 T$

According to Kepler's law of periods
$T^{2} \propto R^{3}$
So, $\Rightarrow \frac{T _{1}}{T _{2}} = (\frac{R _{1}}{R _{2}})^{\frac{3}{2}}$
$\Rightarrow \frac{T}{T _{2}} = (\frac{R}{9 R})^{\frac{3}{2}} = \frac{1}{27} [∵ T_{1} = T]$
$T_{2} = 27 T$

Q. Satellite revolve in radius R has a time period of revolution T. Find time period of a satellite if orbital radius is 9R ?

33​T

9T

27 T

93​T

According to Kepler's law of periods T2∝R3 So,⇒T2​T1​​=(R2​R1​​)23​ ⇒T2​T​=(9RR​)23​=271​[∵T1​=T] T2​=27T

Q. Satellite revolve in radius $R$ has a time period of revolution $T$ . Find time period of a satellite if orbital radius is $9 R$ ?

$33 T$

$93 T$

According to Kepler's law of periods
$T^{2} \propto R^{3}$
So, $\Rightarrow \frac{T _{1}}{T _{2}} = (\frac{R _{1}}{R _{2}})^{\frac{3}{2}}$
$\Rightarrow \frac{T}{T _{2}} = (\frac{R}{9 R})^{\frac{3}{2}} = \frac{1}{27} [∵ T_{1} = T]$
$T_{2} = 27 T$