Chapter 7, Problem 66A

To determine

The height of a rocket that is required so that its weight is half of what it is on Earth.

Answer to Problem 66A

  $h_{requried}=\left(\sqrt{2}-1\right)R$

Explanation of Solution

Given:

The required weight of a rocket is to be half of what it is on Earth.

Formula used:

The weight of an object of mass m on the planet can be obtained by:

  $W=mg$

Where, g is the gravitational field strength on the planet.

And,

The value of g on any planet can be obtained by:

  $g=\frac{GM}{R^2}$

Where, $G$ is the gravitational constant, $R$ is the radius of the planet, and $M$ is the mass of the planet.

Calculation:

As can be seen that weight of an object is directly proportional to the gravitational field, thus to have the weight exactly half of what it is on Earth, the gravitational field is to be exactly half of what it is on Earth’s surface.

Now, using the formula of gravitational field on Earth’s surface,

  $g=\frac{GM}{R^2}\mathrm{....................}\left(1\right)$

If the required height of rocket be “ h ”, then gravitational field at that height can be written as,

  $\frac{g}{2}=\frac{GM}{{\left(R+h\right)}^2}$

Using equation (1) in above,

  $\begin{array}{c}\frac{g}{2}=\frac{g{R}^2}{{\left(R+h\right)}^2}\\ {\left(R+h\right)}^2=2R^2\\ R+h=R\sqrt{2}\\ h=\left(\sqrt{2}-1\right)R\end{array}$

Conclusion:

Thus, the required height is $\left(\sqrt{2}-1\right)$ times the radius of Earth.