The time period of a geostationary satellite is 24 h, at a height 6RE (RE is radius of earth) from surface of earth. The time period of another satellite whose height is 2.5 RE from surface will be,

Question

The time period of a geostationary satellite is 24 h, at a height 6RE (RE is radius of earth) from surface of earth. The time period of another satellite whose height is 2.5 RE from surface will be, (A) 6 √ 2 h (B) 12 √ 2 h (C) (24/2.5) h (D) (12/2.5) h

Tardigrade Admin · Accepted Answer

Keplerâs Third Law :- T propto r3/2 (T2/T1) = ((r2/r1))3/2 =((R + 2.5R/R + 6R))3/2 = (1/2√2) ⇒ T2 = (24/2√2 )= 6√2 hours

Q. The time period of a geostationary satellite is 24 h, at a height 6R_E (R_E is radius of earth) from surface of earth. The time period of another satellite whose height is 2.5 R_E from surface will be,

6 $2$ h

12 $2$ h

$\frac{24}{2.5} h$

$\frac{12}{2.5} h$

Kepler’s Third Law :-
$T \propto r^{3/2}$
$\frac{T _{2}}{T _{1}} = (\frac{r _{2}}{r _{1}})^{3/2}$
$= (\frac{R + 2.5 R}{R + 6 R})^{3/2} = \frac{1}{2 2}$
$\Rightarrow T_{2} = \frac{24}{2 2} = 62$ hours

Q. The time period of a geostationary satellite is 24 h, at a height 6RE (RE is radius of earth) from surface of earth. The time period of another satellite whose height is 2.5 RE from surface will be,

6 2​ h

12 2​ h

2.524​h

2.512​h