91影视

From Newton鈥檚 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鈥檚 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鈥檚 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.,

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

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

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

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

On substituting the given values, you get:

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

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