The time period of a satellite of earth is $$24$$ hours. If the separation between the earth and the satellite is decreased to one fourth of the previous value, then its new time period will become.

Solution :

Time period of satellite is related to orbital radius by Kepler’s third law :

$$T^2 \propto r^3$$

Therefore,

$$\left(\frac{T_2}{T_1}\right)^2 = \left(\frac{r_2}{r_1}\right)^3$$

Given :

$$T_1 = 24\text{ hours}$$

New separation :

$$r_2 = \frac{r_1}{4}$$

Substituting :

$$\left(\frac{T_2}{24}\right)^2 = \left(\frac{1}{4}\right)^3$$

$$\left(\frac{T_2}{24}\right)^2 = \frac{1}{64}$$

$$\frac{T_2}{24} = \frac{1}{8}$$

$$T_2 = \frac{24}{8}$$

$$= 3\text{ hours}$$

Final Answer :

$$3\text{ hours}$$