Calculate the force of Earth’s gravity on a spacecraft 2.00 Earth radii above the Earth’s surface if its mass is 1850 kg.

From Newton’s law of universal gravitation, the attractive force shows an inverse relationship with the square of the distance between objects.And it establishes a direct relationship with the product of the object’s masses.

The magnitude of the gravitational force is given by:

$F_{\mathrm{G}}=\frac{G{m}_1{m}_2}{r^2}$ ... (i)

Here, $G$ is the universal gravitational constant, $m_1$ and $m_2$are the masses of the objects and r is the distance between the objects.

• The mass of the spacecraft is, $m=1850\mathrm{kg}$.
• The distance of the spacecraft from the surface of the Earth is $2r_{\mathrm{E}}$.
• The distance of the spacecraft from the centre of the Earth is, $r=r_{\mathrm{E}}+2r_{\mathrm{E}}=3r_{\mathrm{E}}$.

The force of Earth’s gravity on any object at the surface of the Earth is:

$F_{\mathrm{G}}=mg$

Also, it can be seen from (i) that the force of gravity is inversely proportional to the square of the distance between the spacecraft and the Earth, i.e.,

$\begin{array}{c}{F}_{\mathrm{G}}\propto \frac{1}{r^2}\\ F_{\mathrm{G}}\propto \frac{1}{{\left(3r_{\mathrm{E}}\right)}^2}\\ F_{\mathrm{G}}\propto \frac{1}{9r_{\mathrm{E}}^2}\end{array}$

Here, $r_{\mathrm{E}}$is the radius of the Earth.

Thus, the force of the Earth’s gravity on the spacecraft will be:

$F_{\mathrm{G}}=\frac{mg}{9}$

On substituting the given values, you get:

$\begin{array}{c}{F}_{\mathrm{G}}=\frac{\left(1850\mathrm{kg}\right)\times \left(9.8\mathrm{m}/{\mathrm{s}}^2\right)}{9}\left(\frac{1\mathrm{N}}{1\mathrm{kg}·\mathrm{m}/{\mathrm{s}}^2}\right)\\ =2014.4\mathrm{N}\end{array}$

Therefore, the force of Earth's gravity on the spacecraft is 2014.4 N.