Question

Two satellites A and B revolve around a plant in two coplanar circular orbits in the same sense with radii 10^(4) km and 2 xx 10^(4) km respectively. Time period of A is 28 hours. What is time period of another satellite?

Answer 1

Answer:56√2 hours

Explanation:A/C Kepler's 3rd law of planetary motion T^2=KR^3.

Answer 2

Given :

Time period of satellite A = 28 hours

 R1 =$10^{4}$ km    and R2 = 2 × $10^{4}$ km

To find :

Time period of satellite B

Solution:

Keplar's laws of planetary motion

The 3rd law of keplar says that the square of the orbit's time period of a planet is directly propotional to the cube of the semi-major axis of its orbit.  

$T^{2}$  ∝  $r^{3}$  ;

Where T is the time period and r is the radius of orbit.

R1 = $10^{4}$ km

R2 = 2 ×$10^{4}$ km

$T1^{2}$÷$T2^{2}$= $R1^{3}$÷$R2^3$

$T2^{2}$ = $T1^{2}$ ×  $R2^{3}$ ÷$R1^{3}$

Substitute the values :

T1 = 28 hours R1 = 10000km and R2 = 20000km

T2 = 79.19 hours

Time period of satellite B is 79.19 hours.