Question

The time period of a satellite of earth is $5hr$. If the separation between the earth and the satellite is increased by $3times$ the previous value, the new time period will become

• A. $10hr$
• B. $80hr$
• C. $40hr$
• D. $20hr$

Answer 1

The correct option is C $40hr$

According to Kepler's II law, $T^2\alpha R^3$
$=(\frac{T_1}{T_2}{)}^2=(\frac{R_1}{R_2}{)}^3$
$=(\frac{5}{T_2}{)}^2=(\frac{R}{4R}{)}^3=\frac{1}{64}$
$=\frac{5}{T_2}=\frac{1}{8}=T_2=40hr$