An artificial satellite is moving in a circular orbit around the Earth with a speed equal to half th

Q: Why does the orbital velocity formula contain $R+h$ while the escape velocity formula contains only $R$?

Orbital velocity depends on the actual radius of the circular path, which is the distance from the center of the Earth to the satellite ($R+h$). Escape velocity is defined from the point of launch, which in this case is the surface of the Earth (distance $R$).

Q: Why is the initial kinetic energy zero in part (b)?

The problem states that the satellite is 'stopped suddenly' in its orbit before it begins to fall. Stopping it means its velocity is instantaneously reduced to zero, so its initial kinetic energy is zero.

Q: Does the mass of the satellite affect the final impact speed?

No, as seen in the energy conservation equation, the mass of the satellite $m$ cancels out from both sides. The final speed depends only on the Earth's mass, its radius, and the height from which it falls.

Q: Can we use the formula $v^2 = u^2 + 2gh$ directly for part (b)?

No! The acceleration due to gravity $g$ is not constant over large distances like $h = R$. It varies inversely with the square of the distance from the center of the Earth. Therefore, you must use gravitational potential energy conservation instead of constant-acceleration kinematics.

Q: What is the physical significance of the impact speed being $\sqrt{gR}$?

The speed $\sqrt{gR}$ is exactly equal to the orbital speed of a satellite orbiting very close to the Earth's surface. It is also equal to the escape velocity divided by $\sqrt{2}$.

Question

An artificial satellite is moving in a circular orbit around the Earth with a speed equal to half the magnitude of escape velocity from the Earth. (1990)

(a) Determine the height of the satellite above the earth's surface.

(b) If the satellite is stopped suddenly in its orbit and allowed to fall freely onto the Earth, find the speed with which it hits the surface of the Earth.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine a satellite orbiting the Earth in a stable circular path.

This satellite is bound to the Earth by the invisible but powerful thread of gravity.

Its orbital speed is determined by a delicate balance between the gravitational pull of the Earth and the centripetal acceleration required to keep it in its circular path.

In this problem, we are given a fascinating condition: the orbital speed of this satellite is exactly **half** of the escape velocity from the Earth's surface.

Our goal is to find the height of this satellite above the Earth's surface and then determine its crash speed if it were to be stopped suddenly and allowed to fall freely.

### Part (a): Finding the Height of the Satellite

Let us write down the fundamental equations governing this motion.

The orbital velocity $v_o$ of a satellite at a distance $r = R + h$ from the center of the Earth is given by:

$$v_o = \sqrt{\frac{GM}{R+h}}$$

where $G$ is the universal gravitational constant, $M$ is the mass of the Earth, $R$ is the radius of the Earth, and $h$ is the height of the satellite above the surface.

On the other hand, the escape velocity $v_e$ from the surface of the Earth is the minimum speed required for any object to escape the Earth's gravitational field completely.

It is given by:

$$v_e = \sqrt{\frac{2GM}{R}}$$

According to the problem, the orbital speed is half of the escape velocity:

$$v_o = \frac{v_e}{2}$$

Substituting our expressions into this relation, we get:

$$\sqrt{\frac{GM}{R+h}} = \frac{1}{2} \sqrt{\frac{2GM}{R}}$$

To solve for $h$, we square both sides of the equation to eliminate the square roots:

$$\frac{GM}{R+h} = \frac{1}{4} \left(\frac{2GM}{R}ight)$$

Notice how the term $GM$ cancels out beautifully from both sides.

This simplifies our equation to:

$$\frac{1}{R+h} = \frac{1}{2R}$$

Cross-multiplying gives:

$$2R = R + h \implies h = R$$

This is an incredibly clean and elegant result!

The height of the satellite above the Earth's surface is exactly equal to the radius of the Earth itself.

Since the radius of the Earth $R$ is approximately $6400	ext{ km}$, the height of the satellite is:

$$\mathbf{h = 6400	ext{ km}}$$

### Part (b): The Free Fall Scenario

Now, let us imagine a dramatic turn of events.

The satellite is suddenly stopped in its orbit.

Its orbital velocity instantly drops to zero, and it loses the centripetal support that kept it in orbit.

It begins to fall straight down towards the Earth under the sole influence of gravity.

Since gravity is a conservative force, we can use the **Law of Conservation of Mechanical Energy** to find the speed with which it hits the Earth's surface.

Let the initial state be the point where the satellite is stopped at a distance $r_i = R + h = 2R$ from the center of the Earth.

At this point, its kinetic energy $K_i$ is zero because it has been stopped:

$$K_i = 0$$

Its initial potential energy $U_i$ is:

$$U_i = -\frac{GMm}{2R}$$

Let the final state be the moment of impact on the Earth's surface, where its distance from the center is $r_f = R$.

At this point, its kinetic energy $K_f$ is:

$$K_f = \frac{1}{2}mv^2$$

where $v$ is the impact velocity we want to find.

Its final potential energy $U_f$ is:

$$U_f = -\frac{GMm}{R}$$

By conservation of energy:

$$K_i + U_i = K_f + U_f$$

$$0 - \frac{GMm}{2R} = \frac{1}{2}mv^2 - \frac{GMm}{R}$$

Rearranging the terms to solve for kinetic energy:

$$\frac{1}{2}mv^2 = \frac{GMm}{R} - \frac{GMm}{2R}$$

$$\frac{1}{2}mv^2 = \frac{GMm}{2R}$$

Notice that the mass of the satellite $m$ cancels out from both sides.

This means the crash speed is completely independent of how heavy the satellite is!

Simplifying further:

$$v^2 = \frac{GM}{R}$$

We know that the acceleration due to gravity at the Earth's surface is $g = \frac{GM}{R^2}$, which means $gR = \frac{GM}{R}$.

Substituting this in, we get:

$$v = \sqrt{gR}$$

This is a beautiful and simple formula for the impact speed!

### Final Calculation

Let us plug in the numerical values:

- $g = 9.8	ext{ m/s}^2$
- $R = 6400	ext{ km} = 6.4 	imes 10^6	ext{ m}$

$$v = \sqrt{9.8 	imes 6.4 	imes 10^6}$$

$$v = \sqrt{62.72 	imes 10^6} \approx 7.92 	imes 10^3	ext{ m/s}$$

Converting this to kilometers per second:

$$\mathbf{v \approx 7.9	ext{ km/s}}$$

This is the final speed with which the satellite will strike the Earth's surface.

It is a massive speed, highlighting the immense energy stored in gravitational fields!

Visualized Solution

Visualizing the Satellite's Orbit

Formulas for Orbital and Escape Velocity

Applying the Given Condition

Solving for Height $h$

Calculating the Numerical Value of $h$

Scenario of Free Fall

Setting up the Energy Equation

Solving for Impact Velocity $v$

Calculating the Numerical Value of $v$

The Sigma Insight: Orbital Motion of a Satellite

Analyzing the Setup

Part (a)

Part (b)

Final Calculation

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