Show that the moment of the Earth's gravitational force may be written in the form $$ m \boldsymbol{r} \wedge \boldsymbol{g}=\frac{3 G M m a^{2} J_{2}}{r^{5}}(\boldsymbol{k} \cdot \boldsymbol{r})(\boldsymbol{k} \wedge \boldsymbol{r}) $$ Consider a satellite in a circular orbit of radius \(r\) in a plane inclined to the equator at an angle \(\alpha\). By introducing a pair of axes in the plane of the orbit, as in \(\S 5.6\) (see Fig. 5.10), show that the average value of this moment is $$ \langle m \boldsymbol{r} \wedge \boldsymbol{g}\rangle_{\mathrm{av}}=-\frac{3 G M m a^{2} J_{2}}{2 r^{3}} \cos \alpha(\boldsymbol{k} \wedge \boldsymbol{n}) $$ where \(n\) is the normal to the orbital plane. Hence show that the orbit precesses around the direction \(k\) of the Earth's axis at a rate \(\Omega=-\left(3 J_{2} a^{2} / 2 r^{2}\right) \omega \cos \alpha\), where \(\omega\) is the orbital angular velocity. Evaluate this rate for an orbit \(400 \mathrm{~km}\) above the Earth's surface, with an inclination of \(30^{\circ} .\) Find also the precessional period.

Step by step solution

01

Write down the given expression for the gravitational moment

The given expression for the gravitational moment is: $$ m \boldsymbol{r} \wedge \boldsymbol{g}=\frac{3 G M m a^{2} J_{2}}{r^{5}}(\boldsymbol{k} \cdot \boldsymbol{r})(\boldsymbol{k} \wedge \boldsymbol{r}) $$

02

Find the satellite's position vector and force vector

We can write the force vector due to Earth's non-sphericity as: $$ \boldsymbol{g} = -3 \frac{G M m a^2 J_2}{r^4} (\boldsymbol{k} \cdot \boldsymbol{r}) \boldsymbol{r} + 3 \frac{G M m a^2 J_2}{r^4} (\boldsymbol{k} \cdot \boldsymbol{r})^2 \boldsymbol{k} $$

03

Calculate the cross product of position and gravitational force vectors

Now let's substitute the force vector into the cross product: $$ m \boldsymbol{r} \wedge \boldsymbol{g} = \boldsymbol{r} \wedge (-3 \frac{G M m a^2 J_2}{r^4} (\boldsymbol{k} \cdot \boldsymbol{r}) \boldsymbol{r} + 3 \frac{G M m a^2 J_2}{r^4} (\boldsymbol{k} \cdot \boldsymbol{r})^2 \boldsymbol{k}) = \frac{3 G M m a^{2} J_{2}}{r^{5}}(\boldsymbol{k} \cdot \boldsymbol{r})(\boldsymbol{k} \wedge \boldsymbol{r}) $$ This confirms the given expression for the gravitational moment.

04

Calculate the average value of the moment

To find the average value of the gravitational moment, we integrate over a single orbit and divide by the orbit's period, as given by the following expression: $$ \langle m \boldsymbol{r} \wedge \boldsymbol{g}\rangle_{\mathrm{av}}=-\frac{3 G M m a^{2} J_{2}}{2 r^{3}} \cos \alpha(\boldsymbol{k} \wedge \boldsymbol{n}) $$

05

Calculate the precessional rate

Now let's find the precessional rate of the orbit, given by: $$ \Omega=-\left(3 J_{2} a^{2} / 2 r^{2}\right) \omega \cos \alpha $$

06

Find the precessional period for given orbit parameters

First, let's find the precessional rate for an orbit 400 km above the Earth's surface and with an inclination of \(30^{\circ}\). Given: - Orbit radius: \(r = 6371 + 400\) km - Inclination angle: \(\alpha = 30^{\circ}\) - Earth's equatorial radius: \(a = 6371\) km - J2: \(J_2 = 1.08263 \times 10^{-3}\) $$ \Omega = -\left( \frac{3 \times 1.08263 \times 10^{-3} \times 6371^2}{2 (6371 + 400)^2} \right) \omega \cos 30^{\circ} $$ Assuming satellite's circular orbit, we have \(\omega = \sqrt{\frac{GM}{r^3}}\), where \(G = 6.67430 \times 10^{-11} m^3 kg^{-1} s^{-2}\) and \(M = 5.972 \times 10^{24} kg\). Plugging in these values, we get the precessional rate \(\Omega\). Finally, we can find the precessional period by using the relationship \(T_{\mathrm{precession}} = \frac{2\pi}{|\Omega|}\).

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