Chapter 7, Problem 1STP

To determine

To Choose: The correct option.

Answer to Problem 1STP

Option C

Explanation of Solution

Given:

Orbital radius of satellite 1, $r_1=8.0\times {10}^6\text{m}$

Orbital period of satellite 1, $T_1=1.0\times {10}^6\text{s}$

Orbital radius of satellite 2, $r_2=2.0\times {10}^7\text{m}$

Formula used:

From Kepler’s third law of planetary motion:

  $T^2\propto r^3$

Where, T is the orbital period and r is the orbital radius.

Calculations:

Let orbital period of satellite 2 be $T_2$ .

  $\begin{array}{l}\frac{T_2{}^2}{T_1{}^2}=\frac{r_2{}^3}{r_1{}^3}\\ T_2{}^2=T_1{}^2\left(\frac{r_2{}^3}{r_1{}^3}\right)\\ T_2=\left(1.0\times {10}^6\right){\left(\frac{2.0\times {10}^7}{8.0\times {10}^6}\right)}^{3/2}\\ T_2\approx \text{4}\text{.0}\times {\text{10}}^6\text{s}\end{array}$

Conclusion:

Thus, the option C is correct.