You are explaining to friends why an astronaut feels weightless orbiting in the space shuttle, and they respond that they thought gravity was just a lot weaker up there. Convince them that it isn’t so by calculating how much weaker (in %) gravity is 380 km above the Earth’s surface.

The distance above the Earth’s surface is $r=380\mathrm{km}$.

In order to calculate the distance from the centre of the planet, the radius of the Earth will be added with the given distance above the Earth’s surface.

The relation of acceleration due to gravityis given by,

$\begin{array}{l}g=G\frac{m}{R^2}\\ g=G\frac{m}{{\left(r_E+r\right)}^2}\end{array}$

Here, G is the gravitational constant, m is the mass of the Earth, $r_E$is the radius of the Earth and R is the distance from the centre of the planet.

On plugging the values in the above relation.

$\begin{array}{l}g=\left(6.67\times {10}^{-11}\mathrm{N}·{\mathrm{m}}^2/{\mathrm{kg}}^2\right)\frac{\left(5.98\times {10}^{24}\mathrm{kg}\right)}{{\left(6.38\times {10}^6\mathrm{m}+380\mathrm{km}\times \frac{1000\mathrm{m}}{1\mathrm{m}}\right)}^2}\\ g=8.74\mathrm{m}/{\mathrm{s}}^2\end{array}$

The percentage of acceleration due to gravity in terms of Earth’s acceleration due to gravity $\left(g_E\right)$is calculated as:

$\begin{array}{l}g=\left(8.74\mathrm{m}/{\mathrm{s}}^2\times \frac{1g_E}{9.80\mathrm{m}/{\mathrm{s}}^2}\right)\\ g=0.89g_E\end{array}$

Thus, $g=0.89g_E$is the required value of acceleration due to gravity.