Question 242: Gravitation

Question

Two satellites revolve around a planet in coplanar circular orbits in anticlockwise direction. Their period of revolutions are 1 hour and 8 hours respectively. The radius of the orbit of nearer satellite is 2 $$	imes$$ 10
3
km. The angular speed of the farther satellite as observed from the nearer satellite at the instant when both the satellites are closest is $${\pi  \over x}rad\,{h^{ - 1}}$$ where x is ____________.

Accepted Answer

T
1
= 1 hour
$$\Rightarrow$$ $$\omega$$
1
= 2$$\pi$$ rad/hour
T
2
= 8 hours
$$\Rightarrow$$ $$\omega$$
2
= $${\pi  \over 4}$$ rad/hour
R
1
= 2 $$	imes$$ 10
3
km
As T
2
$$\propto$$ R
3
$$ \Rightarrow {\left( {{{{R_2}} \over {{R_1}}}} ight)^3} = {\left( {{{{T_2}} \over {{T_1}}}} ight)^2}$$
$$ \Rightarrow {{{R_2}} \over {{R_1}}} = {\left( {{8 \over 1}} ight)^{2/3}} = 4 \Rightarrow {R_2} = 8 	imes {10^3}$$ km
V
1
= $$\omega$$
1
R
1
= 4$$\pi$$ $$	imes$$ 10
3
km/h
V
2
= $$\omega$$
2
R
2
= 2$$\pi$$ $$	imes$$ 10
3
km/h
Relative $$\omega$$ = $${{{V_1} - {V_2}} \over {{R_2} - {R_1}}} = {{2\pi  	imes {{10}^3}} \over {6 	imes {{10}^3}}}$$
$$ = {\pi  \over 3}$$ rad/hour
x = 3

Gravitation

Integer / Numeric Answer Type