Chapter 11, Problem 100A

To determine

The mass of earth.

Answer to Problem 100A

  $M=6.031\times {10}^{24}\text{ kg}$

Explanation of Solution

Radius of orbit, $R={10}^7\text{ m}$

Period of revolution of satellite, $T=9.9\times {10}^3\text{ s}$

Formula used:

Orbital velocity of satellite is given by:

  $v=\sqrt{\frac{GM}{R}}$........(1)

Here, G is gravitational constant, M is mass of earth and R is radius.

Velocity is also given by,

  $v=\frac{2\pi R}{T}$........(2)

Here, T is time period and R is radius of orbit.

Calculation:

From equation 1 and equation 2.

  $\sqrt{\frac{GM}{R}}=\frac{2\pi R}{T}$

Gravitational constant $G=6.67\times {10}^{-11}{\text{ Nm}}^2/{\text{kg}}^2$

Now, substitute the value of T, R and G.

  $\begin{array}{l}\sqrt{\frac{6.67\times {10}^{-11}\times M}{{10}^7}}=\frac{2\times 3.14\times {10}^7}{9.9\times {10}^3}\\ \sqrt{\frac{6.67\times {10}^{-11}\times M}{{10}^7}}=0.6343\times {10}^4\\ \sqrt{\frac{6.67\times {10}^{-11}\times M}{{10}^7}}=6343.43\end{array}$

Now, squaring both side.

  $\begin{array}{l}\left(\frac{6.67\times {10}^{-11}\times M}{{10}^7}\right)={6343.43}^2\\ \left(\frac{6.67\times {10}^{-11}\times M}{{10}^7}\right)=4.023\times {10}^7\\ 6.67\times {10}^{-11}M=4.023\times {10}^{14}\\ M=0.6031\times {10}^{25}\\ M=6.031\times {10}^{24}\text{ kg}\end{array}$

Conclusion:

Therefore, mass of earth is $6.031\times {10}^{24}\text{ kg}$ .