Question

At what height above the surface of Earth is there a 1% difference between the approximate magnitude of gravitational field $\mathbf{9}\mathbf{.}\mathbf{8}\mathbf{}\frac{\mathbf{N}}{\mathbf{kg}\mathbf{}}$and the actual value of gravitational field at that location? This is, at what height y above the Earth’s surface is$\frac{\mathbf{G}{\mathbf{M}}_{\mathbf{e}}}{\mathbf{(}{\mathbf{R}}_{\mathbf{e}}\mathbf{+}\mathbf{y}{\mathbf{)}}^{\mathbf{2}}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{99}\frac{\mathbf{G}{\mathbf{M}}_{\mathbf{e}}}{{{\mathbf{R}}_{\mathbf{e}}}^{\mathbf{2}}}$

Short answer

At a height of 32 Km above the surface of Earth, there is a 1% difference between the approximate magnitude of the gravitational field and the actual magnitude of the gravitational field at that location.

Step-by-step solution

Identification of the given data

The given data can be listed below as,

• Assume, the radius of the earth is,$R_e=6400\hspace{0.33em}km$

Significance of gravitational law of motion

The gravitational force formula defines the magnitude of the force between the two objects. It is also known as Newton’s law of gravitation.

Gravitational force = $\frac{(\mathrm{Gravitational}\mathrm{constant})(\mathrm{mass}\mathrm{of}\mathrm{object}1)(\mathrm{mass}\mathrm{of}\mathrm{object}2)}{(\mathrm{distance}\mathrm{between}\mathrm{the}\mathrm{two}\mathrm{objects})}$

Determination of height above the surface of the earth

The value of G at a height “h” above the earth’s surface from the gravitational law of motion can be expressed as,

$g_h=\frac{G{M}_e}{(R_e+h{)}^2}$

Here, G is the gravitational constant,$M_e$ is the mass of the earth,$R_e$ is the radius of Earth.

Rewriting the above equation,

$$\begin{gathered} g_h=\frac{G{M}_e}{R_e{\left(1+{\displaystyle \frac{h}{R_e}}\right)}^2} \\ {g}_h=\frac{g_s}{{\left(1+\frac{h}{R_e}\right)}^2} \end{gathered}$$

for$h<<R_e$

role="math" localid="1658907029410" $$\begin{gathered} g_h=g_s{\left(1+\frac{h}{R_e}\right)}^{-2} \\ =g_s\left(1-\frac{2h}{R_e}\right) \\ {g}_h=g_s\left(1-\frac{2h}{R_e}\right) \\ {g}_h=g_s\left(1-\frac{1}{100}\right) \\ {g}_h=0.99g_s \end{gathered}$$

Therefore,

$0.99=1-\frac{2h}{R_e}$

Now, $R_e=6400km$

Substitute all the values in the above equation.

$$\begin{gathered} \frac{2h}{R_e}=0.01 \\ h=\frac{0.01\times 6400}{2} \\ h=32km \end{gathered}$$

Thus, at height 32 km above the surface of Earth, there is a 1% difference between the approximate magnitude of gravitational field and actual magnitude of gravitational field at that location