At what distance above the surface of the earth is the acceleration due to the earth’s gravity 0.980 m/s² if the acceleration due to gravity at the surface has magnitude 9.8 m/s²?

• The acceleration due to gravity at the unknown distance is g’=0.980 .
• The acceleration due to gravity at the surface of Earth is g=9.80 .

The relation between g and g’ is,

$g^{\text{'}}=\frac{g}{10}$

The acceleration due gravity above a distance h from the surface of Earth can be expressed as,

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

Here, g^' is the acceleration due to gravity at height h from the surface of Earth M is the mass of the Earth, G is the gravitational constant, R is the radius of Earth.

Substitute the value for g^' in the above equation,

$\frac{g}{10}=\frac{GM}{{\left(R+h\right)}^2}\left(i\right)$

The acceleration due to gravity g can be expressed as,

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

Compare the equations (i) and (ii),

$\begin{array}{l}\frac{GM}{10R^2}=\frac{GM}{{\left(R+h\right)}^2}\\ 10R^2={\left(R+h\right)}^2\end{array}$

Evaluate the value of h from the above equation,

$\begin{array}{c}10R^2={\left(R+h\right)}^2\\ \sqrt{10}R=R+h\\ h=R\left(\sqrt{10}-1\right)\end{array}$

Radius of Earth is$R=6.37\times {10}^6\text{\hspace{0.17em}m}$. Thus, the distance h is calculated as,

$\begin{array}{c}h=\left(6.37\times {10}^6\text{\hspace{0.17em}m}\right)\left(\sqrt{10}-1\right)\\ h=1.38\times {10}^7\text{\hspace{0.17em}m}\end{array}$

Thus, the acceleration due to gravity will be 0.980 m/s² at a distance of 1.38 x 10⁷m above the Earth's surface.