Question

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

Answer 1

Kepler's Law of Period states that, if R is orbital radius and T is time period of a satellite,

\longrightarrow\sf{T^2\propto R^3}⟶T

2

∝R

3

Therefore,

\longrightarrow\sf{\left(\dfrac{T_2}{T_1}\right)^2=\left(\dfrac{R_2}{R_1}\right)^3}⟶(

T

1

T

2

)

2

=(

R

1

R

2

)

3

\longrightarrow\sf{\dfrac{T_2}{T_1}=\left(\dfrac{R_2}{R_1}\right)^{\frac{3}{2}}}⟶

T

1

T

2

=(

R

1

R

2

)

2

3

\longrightarrow\sf{T_2=T_1\left(\dfrac{R_2}{R_1}\right)^{\frac{3}{2}}}⟶T

2

=T

1

(

R

1

R

2

)

2

3

Here,

\sf{R_1=R}R

1

=R

\sf{T_1=4\ hr}T

1

=4 hr

\sf{R_2=3R}R

2

=3R

So time period of second satellite will be,

\longrightarrow\sf{T_2=4\left(\dfrac{3R}{R}\right)^{\frac{3}{2}}}⟶T

2

=4(

R

3R

)

2

3

\longrightarrow\sf{T_2=4\times3^{\frac{3}{2}}\ hr}⟶T

2

=4×3

2

3

hr

\longrightarrow\underline{\underline{\sf{T_2=4\sqrt{27}\ hr}}}⟶

T

2

=4

27

hr

Hence (3) is the answer

Answer 2

Question:-

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

Given:-

We know that Kepler's Third law states that, The square of it's period of revolution around Sun is directly proportional to cube of mean distance of a planet from the Sun. Then,

$\longrightarrow\sf{ {T}^{2} ∝ {r}^{3} }⟶$

$\longrightarrow\sf{ \frac{ {T}^{2} }{ {R}^{3} } = Constant \: K \: .... \: (1)}⟶$

$\longrightarrow\sf{ ( { \frac{T_1}{T_2} )}^{2} = ( { \frac{R_2}{R_1} )}^{3} }⟶$

$\longrightarrow\sf{ \frac{T_2}{T_1} = (\frac{R_2}{R_1} ) {}^{\frac{3}{2} } }⟶$

Thus, we get,

$\star \: \boxed{\sf\green{T_2 = t_1( \frac{R_2}{R_1} ) {}^{ \frac{3}{2} } }}⋆$

Thus, Period of second satellite will be:

$\longrightarrow\sf{t_2 = 4( \frac{3R}{R} ) {}^{ \frac{3}{2} } }⟶$

$\longrightarrow\sf{t_2 = 4 \times 3 {}^{ \frac{3}{2} } }$

$\longrightarrow{\underline{\boxed{\sf{T_2 = 4 \sqrt{27} \: hr}}}}⟶$

Thus,

The same planet will have a period of 4 √27 hours.