Two satellites revolve around a planet in coplanar circular orbits in anti-clockwise direction. Thei

Q: Why do we use Kepler's Third Law here?

Kepler's Third Law relates the time period of a satellite to its orbital radius. Since we know the time periods of both satellites and the radius of one, it's the perfect tool to find the unknown radius.

Q: How do we know the satellites are on the same radial line?

The problem states we need to observe them 'at the instant when both the satellites are closest'. In concentric circular orbits, the minimum distance occurs when they are aligned on the same side of the planet.

Q: Why is the relative angular speed formula $\frac{v_1 - v_2}{R_2 - R_1}$?

Relative angular speed is defined as the component of relative velocity perpendicular to the line joining the objects, divided by the distance between them. Since they are aligned radially, their velocities are already perpendicular to the line joining them.

Q: Does the direction of revolution matter?

Yes! Since both revolve anti-clockwise, their velocity vectors point in the same direction at the closest approach. If they revolved in opposite directions, we would have to add their speeds instead of subtracting.

Q: Why did we calculate linear speeds instead of just subtracting angular speeds?

Relative angular speed is NOT simply $\omega_1 - \omega_2$. It depends on the relative linear velocity and the instantaneous distance between the objects. You must convert to linear speeds first to find the true relative perpendicular velocity.

Question

Two satellites revolve around a planet in coplanar circular orbits in anti-clockwise direction. Their period of revolutions are $1	ext{ h}$ and $8	ext{ h}$, respectively. The radius of the orbit of nearer satellite is $2 	imes 10^3	ext{ km}$. The angular speed of the farther satellite as observed from the nearer satellite at the instant when both the satellites are closest is $\frac{\pi}{x}	ext{ rad h}^{-1}$, where $x$ is ……… .

Sigma Percentile · Accepted Answer

### The Cosmic Setup

Imagine you are an astronomer stationed on a satellite, hurtling through the void of space, tracking another satellite orbiting the same distant planet. How fast does the other satellite appear to move from your perspective? This isn't just a math problem; it's a thrilling lesson in relative kinematics and orbital mechanics.

We are given the time periods of both satellites: $T_1 = 1	ext{ h}$ and $T_2 = 8	ext{ h}$. We also know the radius of the inner orbit is $R_1 = 2000	ext{ km}$. To understand their relative motion, we first need to map out the complete geometry of their orbits.

### Kepler's Elegant Law

To find the missing radius of the outer satellite, we invoke the grand architect of orbital mechanics: Johannes Kepler. His Third Law is a beautiful symphony of space and time, stating that the square of the time period is directly proportional to the cube of the orbital radius ($T^2 \propto R^3$).

Let's set up the ratio to compare the two satellites:

$$ \left(\frac{T_1}{T_2}ight)^2 = \left(\frac{R_1}{R_2}ight)^3 $$

Substituting the known values into this elegant equation:

$$ \left(\frac{1}{8}ight)^2 = \left(\frac{2000}{R_2}ight)^3 $$

$$ \frac{1}{64} = \left(\frac{2000}{R_2}ight)^3 $$

Taking the cube root of both sides simplifies the math beautifully:

$$ \frac{1}{4} = \frac{2000}{R_2} \implies R_2 = 8000	ext{ km} $$

### The Speed of the Dance

Now that we have the dimensions of the orbits, we need to understand how fast these satellites are tearing through space. The angular speed $\omega$ is simply $2\pi$ divided by the time period.

$$ \omega_1 = \frac{2\pi}{1} = 2\pi	ext{ rad/h} $$

$$ \omega_2 = \frac{2\pi}{8} = \frac{\pi}{4}	ext{ rad/h} $$

But angular speed alone isn't enough for relative motion; we need their actual linear speeds. We convert these using the fundamental relation $v = \omega R$:

$$ v_1 = 2\pi 	imes 2000 = 4000\pi	ext{ km/h} $$

$$ v_2 = \frac{\pi}{4} 	imes 8000 = 2000\pi	ext{ km/h} $$

### The Climax: Relative Motion

The most thrilling part of the problem occurs when the satellites are closest to each other. In concentric circular orbits, this happens when they align perfectly on the same radial line extending from the planet. At this exact instant, their velocity vectors are perfectly parallel.

Because they are moving in the same anti-clockwise direction, the relative velocity is the difference between their linear speeds:

$$ v_{rel} = v_1 - v_2 = 4000\pi - 2000\pi = 2000\pi	ext{ km/h} $$

The distance separating them is simply the difference in their orbital radii:

$$ r_{rel} = R_2 - R_1 = 8000 - 2000 = 6000	ext{ km} $$

### The Final Revelation

The relative angular speed is defined as the relative perpendicular velocity divided by the separation distance. Since they are aligned radially, their velocities are already perfectly perpendicular to the line of sight between them.

$$ \omega_{rel} = \frac{v_{rel}}{r_{rel}} = \frac{2000\pi}{6000} = \frac{\pi}{3}	ext{ rad/h} $$

Comparing this elegant result to the given expression $\frac{\pi}{x}$, we can clearly see that **$x = 3$**.

Always remember, in the cosmic dance of relative motion, direction is everything. If one satellite had been moving clockwise, their relative velocity would have been the sum of their speeds, drastically changing the final angular speed!

Enter Numerical Value:

Visualized Solution

$Visual Anchor$

$Kepler’s Third Law$

$Raw Setup$

$Atomic Compute$

$Angular Speeds$

$Linear Speeds$

$Relative Angular Speed$

${\text{Substitute & Execute}}$

$Final Answer$

$The Way Forward$

The Sigma Insight: Orbital Motion of a Satellite

The Cosmic Setup

Kepler's Elegant Law

The Speed of the Dance

The Climax

The Final Revelation

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List-I

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Enter Numerical Value:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Visual Anchor

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Kepler’s Third Law

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Raw Setup

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Atomic Compute

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Angular Speeds

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Linear Speeds

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Relative Angular Speed

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } {\text{Substitute & Execute}}

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Answer

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Way Forward