Question

A geostationary satellite is orbiting the earth at a height of 7R above the surface of the earth, where R is the radius of the earth. What will be the time period of another satellite at a height 5R from the surface of the earth?

• A. 9 hours
• B. 6 hours
• C. $9\sqrt{3}hours$
• D. $6\sqrt{3}$ hours

Answer 1

The correct option is C $9\sqrt{3}hours$

Time period of geostationary satellite, $T_1=24h$

Radius of orbit of the satellite, $R_1=7R+R=8R$

Radius of orbit of new satellite, $R_2=5R+R=6R$

Let the time period of the new satellite be $T_2$

According to Kepler’s third law, $T^2\propto r^3\Rightarrow {\left(\frac{T_2}{T_1}\right)}^2={\left(\frac{R_2}{R_1}\right)}^2\Rightarrow {\left(\frac{T_2}{T_1}\right)}^2={\left(\frac{6R}{8R}\right)}^3\Rightarrow {\left(\frac{T_2}{24}\right)}^2={\left(\frac{6R}{8R}\right)}^3\Rightarrow (T_2{)}^2={\left(\frac{6}{8}\right)}^3\times (24{)}^2\Rightarrow T_2=9\sqrt{3}h$

Hence, the correct answer is option (c).