A geostationary satellite orbits around the earth in a circular orbit of radius 36,000 km. Then, the

Q: Why can we assume the spy satellite's orbital radius is just $R_e$?

The problem states the spy satellite is orbiting 'a few hundred kilometers' above the surface. Compared to the Earth's radius of $6400\text{ km}$, an extra $200\text{ km}$ or $300\text{ km}$ is a very small fraction (about $3\%$ to $5\%$). For an approximation, treating $r \approx R_e$ is highly accurate and keeps the math clean.

Q: Where does the $24\text{ hours}$ for the geostationary satellite come from?

By definition, a geostationary satellite must remain directly above the same point on the Earth's equator. For this to happen, its orbital period must match the Earth's rotational period exactly, which is $24\text{ hours}$.

Q: Why is the minimum orbital period around Earth about $84$ minutes?

This is a fundamental limit set by Earth's gravity and size. If a satellite orbits any faster at that radius, it would fly off into an elliptical or escape trajectory. If it orbits slower, gravity would pull it down to the surface. The balance point yields a period of $2\pi\sqrt{R_e/g} \approx 84.3\text{ minutes}$.

Q: Does the mass of the spy satellite affect its orbital period?

No! In gravitational orbits, the gravitational force (which is proportional to the satellite's mass) is balanced by the centripetal acceleration requirement (also proportional to mass). The mass cancels out completely, meaning a tiny bolt and a massive space station in the same orbit will have the exact same orbital period.

Q: What happens to the period if the spy satellite goes even higher?

According to Kepler's Third Law ($T^2 \propto r^3$), as the orbital radius $r$ increases, the orbital period $T$ must also increase. The satellite moves slower in a larger circle, taking more time to complete one full revolution.

Question

A geostationary satellite orbits around the earth in a circular orbit of radius $36,000	ext{ km}$. Then, the time period of a spy satellite orbiting a few hundred km above the earth's surface ($R_e = 6400	ext{ km}$) will approximately be

Options:

(A) $\frac{1}{2}	ext{ h}$

(B) $1	ext{ h}$

(C) $2	ext{ h}$

(D) $4	ext{ h}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

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

Far out in space, at an altitude of about $36,000	ext{ km}$, sits a **geostationary satellite**.

Because its orbital period matches the Earth's rotation exactly ($24	ext{ hours}$), it appears completely stationary relative to a ground observer.

Now, contrast this with a **spy satellite**.

To capture high-resolution images of the surface, it must orbit as closely as possible—typically just a few hundred kilometers above the ground.

This means its orbital radius is practically equal to the radius of the Earth, $R_e = 6400	ext{ km}$.

Our goal is to find the approximate orbital period of this low-Earth-orbit spy satellite using our knowledge of gravitation.

### The Master Equation: Kepler's Third Law

To relate the orbits of these two satellites, we turn to **Kepler's Third Law of Planetary Motion**.

This fundamental law states that the square of the orbital period $T$ of a satellite is directly proportional to the cube of its orbital radius $r$:

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

By taking the ratio of this relationship for both satellites, we can write a clean comparative equation:

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

Here, $T_1 = 24	ext{ h}$ is the period of the geostationary satellite, and $r_1 = 36,000	ext{ km}$ is its orbital radius.

For the spy satellite, its radius $r_2$ is approximately equal to the Earth's radius, $R_e = 6400	ext{ km}$.

### Calculating the Approximation

Let's first simplify the ratio of the radii:

$$\frac{r_2}{r_1} \approx \frac{6400}{36,000} = \frac{8}{45} \approx 0.178$$

Now, we substitute this back into our ratio equation to solve for $T_2$:

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

$$T_2 \approx 24 	imes (0.178)^{1.5}$$

Calculating $(0.178)^{1.5}$ gives approximately $0.075$.

Multiplying this by $24$ hours:

$$T_2 \approx 24 	imes 0.075 \approx 1.8	ext{ hours}$$

This calculation immediately points us towards **$2	ext{ h}$** as the closest option.

### Alternative Verification: Minimum Orbital Period

We can double-check this result using a beautiful first-principles approach.

What is the absolute minimum time it takes for any object to orbit the Earth without crashing?

This occurs when the satellite is skimming just above the surface, where the gravitational force provides the necessary centripetal force:

$$\frac{G M_e m}{R_e^2} = \frac{m v^2}{R_e} \implies v = \sqrt{\frac{G M_e}{R_e}} = \sqrt{g R_e}$$

The time period for this limiting orbit is:

$$T_{	ext{min}} = \frac{2\pi R_e}{v} = 2\pi \sqrt{\frac{R_e}{g}}$$

Substituting $R_e = 6.4 	imes 10^6	ext{ m}$ and $g = 9.8	ext{ m/s}^2$:

$$T_{	ext{min}} = 2\pi \sqrt{\frac{6.4 	imes 10^6}{9.8}} \approx 5060	ext{ seconds} \approx 84.3	ext{ minutes} \approx 1.4	ext{ hours}$$

Since our spy satellite orbits "a few hundred kilometers" above the surface, its orbital radius is slightly larger than $R_e$.

According to Kepler's law, a larger radius means a longer period.

Therefore, the period of the spy satellite must be strictly greater than $1.4	ext{ hours}$.

Looking at the given options:
- (a) $0.5	ext{ h}$ (Impossible, less than the absolute minimum)
- (b) $1	ext{ h}$ (Impossible, less than the absolute minimum)
- (c) $2	ext{ h}$ (Physically realistic and slightly greater than $1.4	ext{ h}$)
- (d) $4	ext{ h}$ (Too large for a low-Earth orbit)

Thus, the most appropriate and accurate approximation is **$2	ext{ h}$**.

Animated Solution for Physics - Gravitation: A geostationary satellite orbits around the earth in a circular orbit of radius $36, 000 km$ . Then, the time period of a spy satellite orbiting a few hundred km above the earth's surface ( $R_{e} = 6400 km$ ) will approximately be

Select Answer:

Visualized Solution

Visualizing the Two Orbits

Kepler's Third Law

Setting up the Ratio

Substituting the Known Values

Calculating the Radius Ratio

Computing the Spy Satellite's Period

Alternative Method: Minimum Orbital Period

Calculating the Minimum Period

Selecting the Best Approximation

The Sigma Insight: Orbital Motion of a Satellite

Analyzing the Setup

The Master Equation

Calculating the Approximation

Alternative Verification

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