Question

what altitude above Earth’s surface would the gravitational acceleration be$\mathbf{4}\mathbf{.}\mathbf{9}\mathbf{m}\mathbf{/}{\mathbf{s}}^{\mathbf{2}}$?

Short answer

The altitude above the earth’s surface where g is$4.9m/s^2$ is$2.6\times {10}^{6}m$

Step-by-step solution

The given data

The gravitational acceleration$a_g=4.9m/s^2$

Understanding the concept of Gravitational acceleration

We use the concept of gravitational acceleration which is in terms of G, M, and r. We can write the equation for gravity at height h then we can solve it for h.

Formula:

Gravitational Acceleration, $a_g=\frac{GM}{r^2}$ (i)

Calculation of the altitude above the Earth’s surface

We can write equation (i) using height h,

$$\begin{gathered} a_g=\frac{GM}{{\left(R_e+h\right)}^2} \\ {R}_e+h=\sqrt{\frac{GM}{a_g}} \\ h=\sqrt{\frac{GM}{a_g}-R_e}................\left(1\right) \end{gathered}$$

Substituting the values of mass & radius of the Earth in equation (1), we get

$$\begin{gathered} h=\sqrt{\frac{(6.67\times {10}^{-11})(5.97\times \times {10}^{24})}{4.9}}-(6.37\times {10}^6) \\ h=2.6\times {10}^{6}m \end{gathered}$$

The altitude above Earth’s surface is at$2.6\times {10}^{6}m$

Using Newton’s law of gravitation, we can find the height h where value of ‘g’ reduces to half its original value.