Question

A satellite has a mass of 6403 kg and is in a circular orbit 4.42 × 105 m above the surface of a planet. The period of the orbit is 2.3 hours. The radius of the planet is 4.73 × 106 m. What would be the true weight of the satellite if it were at rest on the planet’s surface?

Answer 1

A satellite has a mass of $6403\,\mathrm{kg}$ and is in a circular orbit $4.42 \times 105\,\mathrm{m}$ above the surface of a planet. The period of the orbit is 2.3 hours. The radius of the planet is $4.73 \times 106\,\mathrm{m}$. What would be the true weight of the satellite if it were at rest on the planet's surface?

Solution

The height of orbit $H$. The radius of the planet - $R$. Distance between the planet and the satellite $H + R$.

The centripetal force $F_{c}$ is provided by the force of gravity $F_{g}$:

$$F_{g} = F_{c} \rightarrow \frac{GM\,m}{(H + R)^{2}} = m\omega^{2}(H + R),$$

where $G$ – constant of gravity, $M$ – the mass of the planet, $m$ – mass of satellite, $\omega$ – angular velocity of the orbit.

We know that $\omega = \frac{2\pi}{T}$, where $T$ – the period of the orbit.

Now we have

$$\frac{GM\,m}{(H + R)^{2}} = \frac{4\pi^{2}m(H + R)}{T^{2}} \rightarrow \frac{GM}{(H + R)^{2}} = \frac{4\pi^{2}(H + R)}{T^{2}}.$$

The mass of the planet

$$M = \frac{4\pi^{2}((H + R)^{3})}{T^{2}G} = \frac{4\pi^{2}(4.42 * 10^{5} + 4.73 * 10^{6})^{3}}{\left(2.3\,\mathrm{h} * 3600\,\frac{\mathrm{s}}{\mathrm{h}}\right)^{2} * 6.67 * 10^{-11}} = 1.19 * 10^{24}\,\mathrm{kg}.$$

The gravitational force, $F_{g}$, is also equal to $m * g$. To find the acceleration,

$$m\,g = \frac{GM\,m}{R^{2}} \rightarrow g = \frac{GM}{R^{2}}.$$

At the surface of the planet,

$$g = \frac{6.67 * 10^{-11} * 1.19 * 10^{24}}{(4.73 * 10^{6})^{2}} = 3.55\,\frac{\mathrm{m}}{\mathrm{s}^{2}}.$$

So,

$$F_{g} = m\,g = 6403 * 3.55 = 23\,\mathrm{kN}.$$

Answer: 23 kN.