Question

The radius of Earth is 6.371 × 103 km.Find the altitude above Earth’s surface where Earth’s gravitational field strength would be two-thirds of its value at the surface.Find the altitude above Earth’s surface where Earth’s gravitational field strength would be one-fourth of its value at the surface.

Answer 1

Given,

The radius of the earth, r=6.371×10³ km=6.371×10⁶ m

The gravitational strength at the surface of the earth is given by,

$g=\frac{GM}{r^2}_{}$

Where G is the gravitational constant and M is the mass of the earth.

Thus the gravitational field strength at a height h, where the gravitational field strength would be two-third of its values at the surface is,

$g_h=\frac{GM}{(r+h)^2}=\frac{2g}{3}$

Thus,

$\begin{gathered} \frac{GM}{(r+h)^2}=\frac{2GM}{3r^2} \\ \Rightarrow\frac{3}{2}r^2=(r+h)^2 \\ \Rightarrow\sqrt[]{\frac{3}{2}}r=r+h \\ \Rightarrow h=r(\sqrt[]{\frac{3}{2}}-1) \end{gathered}$

On substituting the value of r,

$\begin{gathered} h=6.371\times10^6(\sqrt[]{\frac{3}{2}}-1) \\ =6.371\times10^6(0.225) \\ =1.433\times10^6\text{ m} \end{gathered}$

Thus the earth's gravitational field strength would be two-thirds of its value at the surface at a height of 1.433×10⁶ m above the surface.

The height at which the gravitational field strength would be one-fourth its value at the surface is given by,

$g_H=\frac{GM}{(r+H)^2}=\frac{g}{4}$

Thus,

$\begin{gathered} \frac{GM}{(r+H)^2}=\frac{GM}{4r^2} \\ 4r^2=(r+H)^2 \\ \sqrt[]{4}r=r+H \\ \Rightarrow H=2r-r=r \end{gathered}$

Therefore,

$H=6.371\times10^6\text{ m}$

Thus the earth's gravitational field strength would be one-fourth of its value at the surface at a height of 6.371×10⁶ m above the surface.