A geostationary satellite is orbiting the earth at a height of 6R above the surface of the earth whe

Q: Why did we take the time period of the first satellite as 24 hours? The question doesn't mention 24 hours anywhere!

This is an implicit property of a **geostationary satellite**. By definition, a geostationary satellite remains directly above a fixed point on the Earth's equator. For this to happen, its orbital period must match the Earth's rotational period, which is exactly 24 hours.

Q: I used $r = 6R$ and $r = 2.5R$ and got a completely different answer. What did I do wrong?

This is the most common mistake! The values $6R$ and $2.5R$ are the **heights** ($h$) above the Earth's surface. Gravitational force acts from the center of the Earth, so you must add the Earth's radius $R$ to get the true orbital radius: $r = R + h$.

Q: How do we simplify $2^{3/2}$ without a calculator?

You can break down the fractional exponent: $2^{3/2} = 2^{1 + 1/2} = 2^1 \times 2^{1/2} = 2\sqrt{2}$. Since $\sqrt{2} \approx 1.414$, you can easily compute $2\sqrt{2} \approx 2.828$ or simplify $\frac{12}{\sqrt{2}}$ directly to $6\sqrt{2}$.

Q: Does Kepler's Third Law only apply to planets orbiting the Sun?

No! Kepler's Third Law applies to **any** system of bodies orbiting a common central mass under the influence of gravity, including moons orbiting a planet or artificial satellites orbiting the Earth.

Q: What would happen if the second satellite's height was $0$ (orbiting right at the surface)?

If $h = 0$, then $r = R$. Using Kepler's Third Law, its time period would be $T = 24 \times \left(\frac{R}{7R}\right)^{3/2} = \frac{24}{7\sqrt{7}} \approx 1.4\text{ hours}$ (or about 84 minutes). This is the minimum possible orbital period for any satellite orbiting the Earth!

Question

A geostationary satellite is orbiting the earth at a height of $6R$ above the surface of the earth where $R$ is the radius of earth. The time period of another satellite at a height of $2.5 R$ from the surface of the earth is _________ hours.

Sigma Percentile · Accepted Answer

### Introduction to Satellite Motion

Imagine standing on the surface of the Earth, looking up at the night sky.

High above us, hundreds of artificial satellites are silently gliding through the vacuum of space, locked in a delicate gravitational dance with our planet.

Among these, **geostationary satellites** hold a special place.

They orbit at a very specific altitude such that they take exactly **24 hours** to complete one full revolution, matching the Earth's rotation perfectly.

To an observer on the ground, they appear completely stationary in the sky.

But what happens if we place another satellite in a lower orbit, closer to the Earth?

How does its orbital period change?

This classic JEE problem invites us to explore the beautiful relationship between a satellite's orbital altitude and its time period of revolution using **Kepler's Third Law of Planetary Motion**.

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### The Crucial Definition of Orbital Radius

Before diving into any calculations, we must address the most common pitfall in gravitation problems: **the definition of orbital radius**.

When a problem states that a satellite is at a "height" $h$ above the surface of the Earth, this is **not** its orbital radius.

Gravity acts from the center of mass of the Earth.

Therefore, the orbital radius $r$ must always be measured from the **center of the Earth**.

If the Earth has a radius $R$, then the total orbital radius is:

$$r = R + h$$

Let's apply this fundamental rule to both of our satellites.

#### For the Geostationary Satellite ($S_1$):

The problem states that $S_1$ is orbiting at a height of $6R$ above the surface.

$$r_1 = R + 6R = 7R$$

#### For the Second Satellite ($S_2$):

The second satellite is orbiting at a height of $2.5R$ above the surface.

$$r_2 = R + 2.5R = 3.5R$$

Now we have the true orbital radii of both satellites measured from the center of the Earth.

Notice that the orbital radius of the second satellite is exactly half that of the geostationary satellite:

$$\frac{r_2}{r_1} = \frac{3.5R}{7R} = \frac{1}{2}$$

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### Kepler's Third Law: The Bridge of Proportionality

To find the time period of the second satellite, we need a law that connects the orbital radius $r$ to the time period $T$.

This is precisely what **Kepler's Third Law** (also known as the Law of Harmonies) does.

It states that the square of the orbital period of a satellite is directly proportional to the cube of its semi-major axis (or radius for a circular orbit):

$$T^2 \propto r^3$$

Taking the square root of both sides, we get:

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

This proportionality is incredibly powerful because it allows us to set up a ratio between the two satellites, completely bypassing the need to know the mass of the Earth or the universal gravitational constant $G$:

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

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### Step-by-Step Calculation

Now, let's substitute our known values into this ratio.

$$\frac{T_2}{T_1} = \left(\frac{3.5R}{7R}ight)^{3/2}$$

$$\frac{T_2}{T_1} = \left(\frac{1}{2}ight)^{3/2}$$

To simplify $\left(\frac{1}{2}ight)^{3/2}$, we can write it as:

$$\left(\frac{1}{2}ight)^{3/2} = \frac{1}{2^{3/2}} = \frac{1}{2\sqrt{2}}$$

So, our ratio equation becomes:

$$T_2 = T_1 	imes \frac{1}{2\sqrt{2}}$$

We know that $S_1$ is a **geostationary satellite**, which means its time period $T_1$ is exactly **24 hours**.

Let's substitute $T_1 = 24$ hours into our equation:

$$T_2 = 24 	imes \frac{1}{2\sqrt{2}}$$

$$T_2 = \frac{12}{\sqrt{2}}$$

To rationalize the denominator, we multiply the numerator and denominator by $\sqrt{2}$:

$$T_2 = \frac{12\sqrt{2}}{2} = 6\sqrt{2}	ext{ hours}$$

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### Final Numerical Value

To find the decimal value for our fill-in-the-blank answer, we substitute the approximation $\sqrt{2} \approx 1.414$:

$$T_2 = 6 	imes 1.414 = 8.484	ext{ hours}$$

Rounding to two decimal places, we get:

$$T_2 \approx 8.48	ext{ hours}$$

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### Physical Intuition and Conclusion

Let's take a moment to appreciate the physics of this result.

The second satellite is closer to the Earth ($3.5R$ compared to $7R$).

Because it is closer, the gravitational pull of the Earth is stronger.

To maintain a stable circular orbit without falling into the Earth, the satellite must travel at a **higher orbital speed**.

Combined with a **shorter orbital path** (smaller circumference), a higher speed means it completes its orbit much faster—taking only **8.48 hours** compared to the geostationary satellite's 24 hours.

This beautiful harmony of speed, distance, and time is the essence of orbital mechanics!

Animated Solution for Physics - Gravitation: A geostationary satellite is orbiting the earth at a height of $6 R$ above the surface of the earth where $R$ is the radius of earth. The time period of another satellite at a height of $2.5 R$ from the surface of the earth is _________ hours.

Enter Numerical Value:

Visualized Solution

Visualizing the Satellite Orbits

Defining Orbital Radii from the Center

Orbital Radius of the Geostationary Satellite ( $S_{1}$ )

Orbital Radius of the Second Satellite ( $S_{2}$ )

Applying Kepler's Third Law

Formulating the Ratio of Time Periods

Substituting the Orbital Radii

Simplifying the Fraction

Recall the Time Period of $S_{1}$

Solving for $T_{2}$

Calculating the Decimal Value

Conceptual Takeaway

The Sigma Insight: Orbital Motion of a Satellite

Introduction to Satellite Motion

The Crucial Definition of Orbital Radius

# For the Geostationary Satellite ( $S_{1}$ )

# For the Second Satellite ( $S_{2}$ )

Kepler's Third Law

Step-by-Step Calculation

Final Numerical Value

Physical Intuition and Conclusion

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