91影视

The given data can be listed below as,

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

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

The value of G at a height 鈥渉鈥� above the earth鈥檚 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" $$\begin{gathered} g_h=g_s{\left(1+\frac{h}{R_e}\right)}^{-2} \\ =g_s\left(1-\frac{2h}{R_e}\right) \\ {g}_h=g_s\left(1-\frac{2h}{R_e}\right) \\ {g}_h=g_s\left(1-\frac{1}{100}\right) \\ {g}_h=0.99g_s \end{gathered}$$

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

Now, $R_e=6400km$

Substitute all the values in the above equation.

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

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