Question

A missile which missed its target, went into orbit around the earth. The radius of the orbit is four times the radius of the parking orbit of a satellite. The period of the missile as satellite is –
a) $2$ days
b) $4$ days
c) $8$ days
d) $16$ days

Answer 1

Kepler's law provides the relation between the radius and the time period of the round path. According to Kepler, the square of the time period is directly proportional to the cube of the circular route radius.

Complete step-by-step solution:
Let $R_{p}$ and $T_{p}$ are the radius and time period for parking satellites.
$T_{p} = 1$ day
Let $R_{m}$ and $T_{m}$ are the radius and time period for a missile satellite.
$R_{m} = 4 R_{p}$
We will use Kepler’s third law:
$T^{2} \propto R^{3}$
$\implies T^{2} =k R^{3}$
We will apply the above formula for parking satellites and missile satellites.
$\dfrac{T_{m}^{2}}{T^{2}_{p}} = \dfrac{R_{m}^{3}}{R^{3}_{p}}$
Put all the values of given values in the above formula:
$\dfrac{T_{m}^{2}}{T^{2}_{p}} = \dfrac{64 R_{p}^{3}}{R^{3}_{p}}$
$\dfrac{T_{m}^{2}}{T^{2}_{p}} = 64$
$\implies T_{m} = 8 T_{p}$
Put $T_{p} = 1$ day in the above formula.
$\implies T_{m} = 8 \times 1$
$\implies T_{m} = 8$ days.

Note: The square of a time period of a body orbiting in a circular orbit is directly proportionate to the cube of the radius of the round orbit—the field of circles. Satellites or whatever other celestial body revolving in circular paths of an equal radius have an equal duration of time.