Respuesta :
Answer:
Time period of second satellite will be 50.904 hour
Explanation:
We have given time period of first satellite [tex]T_1=18hour[/tex]
Orbital radius of first satellite [tex]a_1=2\times 10^7m[/tex]
Orbital radius of second satellite [tex]a_2=4\times 10^7m[/tex]
We have to find the time period of second satellite.
Time period of satellite is equal to [tex]T=2\pi \sqrt{\frac{a^3}{G(M_1+M_2)}}[/tex]
From the relation we can see that [tex]T\propto a^{\frac{3}{2}}[/tex]
[tex]\frac{T_1}{T_2}=\frac{a_1^\frac{3}{2}}{a_2^\frac{3}{2}}[/tex]
[tex]\frac{18}{T_2}=\frac{(2\times 10^7)^\frac{3}{2}}{(4\times 10^7)^\frac{3}{2}}[/tex]
[tex]\frac{18}{T_2}=(\frac{1}{2})^\frac{3}{2}[/tex]
[tex]T_2=50.904hour[/tex]
Time period of second satellite will be 50.904 hour
