Two satellites S1 and S2 revolve round a planet in coplanar circular orbits in the same sense. Their

Q: Why is the relative velocity negative in part (a)?

The relative velocity is defined as $\vec{v}_{2/1} = \vec{v}_2 - \vec{v}_1$. Since the inner satellite $S_1$ is moving faster than the outer satellite $S_2$ ($v_1 > v_2$), an observer on $S_1$ will see $S_2$ lagging behind, which physically corresponds to moving backward (hence the negative sign).

Q: Can we use the formula $\omega = v/r$ directly for relative angular velocity?

Yes, but only if the relative velocity is perpendicular to the line joining the two objects. At the point of closest approach, the line joining the satellites is radial and their velocities are tangential, so the relative velocity is perfectly perpendicular, allowing us to use $\omega_{\text{rel}} = \frac{v_{\text{rel}}}{r_{\text{rel}}}$ directly.

Q: What if the satellites were moving in opposite directions?

If they were moving in opposite directions (opposite senses), their relative velocity at closest approach would be the sum of their speeds, i.e., $v_{\text{rel}} = v_2 - (-v_1) = v_2 + v_1$. This would significantly increase both the relative speed and the relative angular velocity.

Q: How did we convert km/h to m/s in the final step?

To convert kilometers per hour to meters per second, we multiply by $\frac{5}{18}$ (since $1\text{ km} = 1000\text{ m}$ and $1\text{ h} = 3600\text{ s}$, giving $\frac{1000}{3600} = \frac{5}{18}$).

Q: Is the angular velocity constant throughout their orbits?

No. The relative angular velocity of $3.0 \times 10^{-4}\text{ rad/s}$ is calculated specifically for the moment of closest approach. As the satellites move along their respective orbits, the distance between them and the angle of their relative velocity vector will change, causing the relative angular velocity to vary continuously.

Question

Two satellites $S_1$ and $S_2$ revolve round a planet in coplanar circular orbits in the same sense. Their periods of revolution are $1	ext{ h}$ and $8	ext{ h}$, respectively. The radius of the orbit of $S_1$ is $10^4	ext{ km}$ when $S_2$ is closest to $S_1$. Find
(a) the speed of $S_2$ relative to $S_1$,
(b) the angular speed of $S_2$ as actually observed by an astronaut in $S_1$.

Sigma Percentile · Accepted Answer

### Introduction to Orbital Dynamics

Imagine standing on a planet, looking up at the night sky, and watching two artificial satellites trace elegant, concentric paths across the heavens.

This problem invites us to step into the shoes of an astronaut aboard one of these satellites, $S_1$, and observe the motion of a neighboring satellite, $S_2$, at their moment of closest approach.

To solve this, we must bridge the gap between absolute orbital motion governed by gravity and the relative kinematics experienced by observers in non-inertial frames.

---

### Analyzing the Setup with Kepler's Laws

We are given two satellites revolving in coplanar circular orbits in the same sense (direction) around a central planet.

Their time periods of revolution are:

$$T_1 = 1	ext{ h}$$

$$T_2 = 8	ext{ h}$$

We are also given that the radius of the orbit of $S_1$ is:

$$r_1 = 10^4	ext{ km}$$

To find the radius of the second satellite's orbit, $r_2$, we invoke **Kepler's Third Law of Planetary Motion**, which states that the square of the orbital period is proportional to the cube of the semi-major axis (or radius for circular orbits):

$$T^2 \propto r^3 \implies r \propto T^{2/3}$$

Using this proportionality, we can set up a ratio between the two orbits:

$$\frac{r_2}{r_1} = \left(\frac{T_2}{T_1}ight)^{2/3}$$

Substituting the known values:

$$r_2 = 10^4 	imes \left(\frac{8}{1}ight)^{2/3} = 10^4 	imes (2^3)^{2/3} = 4 	imes 10^4	ext{ km}$$

This tells us that the second satellite orbits at exactly four times the distance of the first satellite from the planet's center.

---

### Finding the Absolute Orbital Speeds

Now that we have the orbital radii for both satellites, we can compute their absolute speeds.

For a circular orbit, the speed is simply the circumference of the orbit divided by the time period of revolution:

$$v = \frac{2\pi r}{T}$$

Let's calculate this for both satellites:

For $S_1$:

$$v_1 = \frac{2\pi 	imes 10^4	ext{ km}}{1	ext{ h}} = 2\pi 	imes 10^4	ext{ km/h}$$

For $S_2$:

$$v_2 = \frac{2\pi 	imes (4 	imes 10^4	ext{ km})}{8	ext{ h}} = \pi 	imes 10^4	ext{ km/h}$$

Notice that even though $S_2$ has a larger orbit, its speed is exactly half of $S_1$'s speed. This is a fundamental characteristic of gravitational orbits: **the further out you are, the slower you travel**.

---

### Part (a): Relative Velocity at Closest Approach

At the moment of closest approach, the two satellites, $S_1$ and $S_2$, and the center of the planet lie on a single straight line.

Because they are moving in the same sense, their velocity vectors are parallel and point in the same direction.

Therefore, the velocity of $S_2$ relative to $S_1$ is given by the simple vector difference:

$$v_{	ext{rel}} = v_2 - v_1$$

$$v_{	ext{rel}} = \pi 	imes 10^4 - 2\pi 	imes 10^4 = -\pi 	imes 10^4	ext{ km/h}$$

The negative sign indicates that to an astronaut on $S_1$, the outer satellite $S_2$ appears to be drifting backward at a speed of $\pi 	imes 10^4	ext{ km/h}$.

---

### Part (b): Relative Angular Velocity

Now, let's look at the angular speed of $S_2$ as observed by an astronaut in $S_1$.

By definition, the relative angular velocity $\omega_{	ext{rel}}$ of one object with respect to another is the component of their relative velocity perpendicular to the line joining them, divided by the distance between them:

$$\omega_{	ext{rel}} = \frac{v_{	ext{rel}, \perp}}{r_{	ext{rel}}}$$

At the moment of closest approach, the line joining the two satellites is radial.

Since both velocity vectors are tangential, the relative velocity vector is already perpendicular to the line joining them.

Thus, the perpendicular component of relative velocity is simply the magnitude of the relative velocity:

$$v_{	ext{rel}, \perp} = |v_2 - v_1| = \pi 	imes 10^4	ext{ km/h}$$

The distance between the two satellites at closest approach is:

$$r_{	ext{rel}} = r_2 - r_1 = 4 	imes 10^4 - 10^4 = 3 	imes 10^4	ext{ km}$$

Now, we substitute these values into our angular velocity formula. To get the answer in standard SI units (radians per second), we must convert kilometers to meters and hours to seconds:

$$v_{	ext{rel}, \perp} = \pi 	imes 10^4	ext{ km/h} = \pi 	imes 10^4 	imes \frac{1000	ext{ m}}{3600	ext{ s}} = \pi 	imes 10^4 	imes \frac{5}{18}	ext{ m/s}$$

$$r_{	ext{rel}} = 3 	imes 10^4	ext{ km} = 3 	imes 10^7	ext{ m}$$

Now, compute $\omega_{	ext{rel}}$:

$$\omega_{	ext{rel}} = \frac{\pi 	imes 10^4 	imes \frac{5}{18}}{3 	imes 10^7} = \frac{5\pi}{54 	imes 10^3}	ext{ rad/s}$$

$$\omega_{	ext{rel}} \approx \frac{15.708}{54000} \approx 2.91 	imes 10^{-4}	ext{ rad/s} \approx 3.0 	imes 10^{-4}	ext{ rad/s}$$

This extremely small angular speed reflects the vast distances involved in space travel, showing how slowly objects appear to rotate across the cosmic background even when traveling at thousands of kilometers per hour.

Visualized Solution

Visualizing the Coplanar Orbit Setup

Applying Kepler's Third Law

Calculating the Orbital Radius $r_{2}$

Calculating Orbital Speed $v_{1}$

Calculating Orbital Speed $v_{2}$

Part (a): Relative Speed of $S_{2}$ with respect to $S_{1}$

Part (b): Relative Angular Velocity Formula

Computing and Converting $ω_{rel}$ to SI Units

The Sigma Insight: Orbital Motion of a Satellite

Introduction to Orbital Dynamics

Analyzing the Setup with Kepler's Laws

Finding the Absolute Orbital Speeds

Part (a)

Part (b)

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