Question

A satellite GeoSAT is in a circular geostationary orbit of radius $R_{G}$, above a point P on the equator. Another satellite ComSAT is in a lower circular orbit of radius $0.81
R_{G}$ At $7$PM on January $1$ ComSAT is sighted directly above P. On which day among
the following can ComSAT be sighted directly above P between $7$ PM and $8$ PM
A) January $ 3$
B) January $9$
C) January $15$
D) January $21$

Answer 1

Hint: Time period of Geostationary satellite is $24$ hours. In the question,
radius and time period are given. So, we will use Kepler’s third Law which gives relation
between time period and radius.
Formula Used:
T is directly related to the orbit semi-major axis through the formula: -
$T^{2} \propto R^{3}$
Where, T is the time period of satellite
R is the radius.

Complete answer:
Given-
Radius of geostationary satellite = $R_{G}$,
Radius of ComSAT =$0.81 R_{G}$,
Time period of geostationary satellite, $T_{G}$= $24$ hours
We have to find the time period of ComSAT. So, we will use Kepler’s third law given by-
$T^{2} \propto R^{3}$
Where, T is the time period of satellite
R is the radius.
So, arrange this formula for geostationary satellites and ComSAT.
$\dfrac{R_{G}^{3}}{ T_{G}^{2}}= \dfrac{ R_{c}^{3}}{ T_{c}^{2}}$
$T_{c} = \sqrt{\dfrac{ {24^{2} \times 0.81 R_{G}^{3}}}{{ R_{G}^{3}}}}$
$T_{c} =17.496$ hours
So, we need to find the day at which ComSAT be sighted directly above P between $7$PM
and $8$PM
$Hours = 17.496 \times 11 = 192.456 $ hours
Now we will find days by dividing the hours with $24$.
$Days = \dfrac {192.456}{24} = 8.019 $ days
Now, the day from Jan $ 1+ 8.01 =9 $January

So, B is the right answer.

Additional Information:
Geostationary satellites possess the unique quality of remaining
fixed in the same state in the sky as seen from any arranged location on Earth, meaning that
ground-based wires do not need to track them but can continue fixed in one direction.

Note:
The time period of the geostationary satellite is not given. so, we will take same time
period as earth has a time period of one day. Noted that square of time period is directly
proportional to the cube of radius. Time period will increase as the radius. T
should be in hours and for days T should be divided by $24$.