Question

A satellite is launched in a circular orbit of radius $R$ around the earth. A second satellite is launched into an orbit of radius $1.01R$. The period of the second satellite is longer than the first one (approximately) by

• A. $1.5\%$
• B. $0.5\%$
• C. $3\%$
• D. $1\%$
• E. $2\%$

Answer 1

The correct option is A

$1.5\%$

Step 1. Given data:

The orbit radius for the first satellite $=R$

The orbit radius for the second satellite $=1.01R$

Step 2. Finding the percentage:

Assume the time periods for the first and the second satellites as $T_1$ and $T_2$ respectively, then

From the Kepler's third law we have, $T\propto R^{\frac{3}{2}}$

Now, $T_1\propto R^{\frac{3}{2}}$

and, $T_2\propto {\left(1.01R\right)}^{\frac{3}{2}}$

Dividing the above two we get, $\frac{T_2}{T_1}={\left(1.01\right)}^{\frac{3}{2}}$

$\Rightarrow$ $T_2=1.015T_1$

Now, the percentage increase $=\left(1.015T_1-T_1\right)\times 100$

$=1.5\%$

Hence, option (A) is correct.