Question

Three satellites orbit the Earth. The first moves in a circular orbit with a radius R₁, the second in a circular orbit with a radius R₂, where R₂ > R₁. The third satellite follows an elliptical path, touching the smaller circular orbit at its perigee in point P and the larger circular orbit at its apogee in point A.

• $V_{k1}$ as the instantaneous speed on the circular trajectory with radius R₁,
• $V_{k2}$ as the instantaneous speed on the circular trajectory with radius R₂,
• $V_p$ as the instantaneous speed at the perigee of the elliptical trajectory,
• $V_a$ as the instantaneous speed at the apogee of the elliptical trajectory.

$T_1$ is the orbital time period of satellite in radius R₁. $T_2$ is the orbital time period of satellite in radius R₂. $T_3$ is the orbital time period of the satellite in elliptical orbit.

• A. $V_a < V_{k2} < V_{k1} < V_p$
• B. $V_a < V_{k1} < V_p < V_{k2}$
• C. $T_1 = 2\pi \sqrt{\frac{R_1^3}{GM}}$
• D. $T_3 = \pi \sqrt{\frac{(R_1+R_2)^3}{2GM}}$

Answer 1

Answer: (A), (C), (D)

A, C, D

Answer 2

The question asks us to compare the speeds of three satellites orbiting the Earth and to provide expressions for their time periods.

Let $G$ be the gravitational constant and $M$ be the mass of the Earth.

For a satellite in a circular orbit of radius $r$, the speed is given by $v_k = \sqrt{\frac{GM}{r}}$. The period of a circular orbit is $T = \frac{2\pi r}{v_k} = 2\pi r \sqrt{\frac{r}{GM}} = 2\pi \sqrt{\frac{r^3}{GM}}$.

For the first satellite in a circular orbit with radius $R_1$, the speed is $V_{k1} = \sqrt{\frac{GM}{R_1}}$. The time period is $T_1 = 2\pi \sqrt{\frac{R_1^3}{GM}}$.

For the second satellite in a circular orbit with radius $R_2$, the speed is $V_{k2} = \sqrt{\frac{GM}{R_2}}$. Since $R_2 > R_1$, we have $\frac{1}{R_2} < \frac{1}{R_1}$, so $V_{k2} < V_{k1}$.

For the third satellite in an elliptical orbit, the perigee is at $R_1$ and the apogee is at $R_2$. The speed at perigee is $V_p$ and the speed at apogee is $V_a$. The semi-major axis of the elliptical orbit is $a = \frac{R_1 + R_2}{2}$. According to Kepler's third law, the period of the elliptical orbit is $T_3 = 2\pi \sqrt{\frac{a^3}{GM}} = 2\pi \sqrt{\frac{(\frac{R_1 + R_2}{2})^3}{GM}} = 2\pi \sqrt{\frac{(R_1 + R_2)^3}{8GM}} = \pi \sqrt{\frac{(R_1 + R_2)^3}{2GM}}$.

Using conservation of energy and angular momentum for the elliptical orbit:

Energy at perigee = Energy at apogee: $\frac{1}{2}mV_p^2 - \frac{GMm}{R_1} = \frac{1}{2}mV_a^2 - \frac{GMm}{R_2}$

Angular momentum at perigee = Angular momentum at apogee: $mR_1 V_p = mR_2 V_a$, so $V_a = \frac{R_1}{R_2} V_p$.

Substituting $V_a$ into the energy equation:

$\frac{1}{2}V_p^2 - \frac{GM}{R_1} = \frac{1}{2}\left(\frac{R_1}{R_2}V_p\right)^2 - \frac{GM}{R_2}$

$\frac{1}{2}V_p^2 \left(1 - \frac{R_1^2}{R_2^2}\right) = GM \left(\frac{1}{R_1} - \frac{1}{R_2}\right)$

$\frac{1}{2}V_p^2 \left(\frac{R_2^2 - R_1^2}{R_2^2}\right) = GM \left(\frac{R_2 - R_1}{R_1 R_2}\right)$

$V_p^2 = \frac{2GM}{R_1 R_2} \frac{R_2^2}{R_1 + R_2} = \frac{2GM R_2}{R_1 (R_1 + R_2)}$

$V_p = \sqrt{\frac{2GM R_2}{R_1 (R_1 + R_2)}}$

$V_a = \frac{R_1}{R_2} V_p = \frac{R_1}{R_2} \sqrt{\frac{2GM R_2}{R_1 (R_1 + R_2)}} = \sqrt{\frac{2GM R_1}{R_2 (R_1 + R_2)}}$

Comparing speeds:

$V_p^2 = \frac{2GM R_2}{R_1 (R_1 + R_2)}$ and $V_{k1}^2 = \frac{GM}{R_1}$. $\frac{V_p^2}{V_{k1}^2} = \frac{2R_2}{R_1 + R_2}$. Since $R_2 > R_1$, $R_1 + R_2 < 2R_2$, so $\frac{2R_2}{R_1 + R_2} > 1$. Thus $V_p > V_{k1}$.

$V_a^2 = \frac{2GM R_1}{R_2 (R_1 + R_2)}$ and $V_{k2}^2 = \frac{GM}{R_2}$. $\frac{V_a^2}{V_{k2}^2} = \frac{2R_1}{R_1 + R_2}$. Since $R_2 > R_1$, $R_1 + R_2 > 2R_1$, so $\frac{2R_1}{R_1 + R_2} < 1$. Thus $V_a < V_{k2}$.

We already have $V_{k2} < V_{k1}$.

Also, from $R_1 V_p = R_2 V_a$ and $R_2 > R_1$, we have $V_p > V_a$.

Combining the inequalities: $V_a < V_{k2} < V_{k1} < V_p$.

Therefore, options (A), (C), and (D) are correct.