What is the minimum energy required to launch a satellite of mass m from the surface of a planet of

Q: Why is the orbital radius $3R$ and not $2R$?

The altitude of $2R$ is measured from the surface of the planet. To find the orbital radius, you must add the planet's radius $R$, giving a total distance of $3R$ from the center.

Q: Why does the satellite have kinetic energy in the final state?

To remain in a stable circular orbit, the satellite must travel at a specific orbital speed. If it had zero kinetic energy, it would simply fall back to the planet.

Q: Can we just calculate the difference in potential energy?

No, calculating only the potential energy difference would give you the energy required to just lift the satellite to that height. It wouldn't have the speed to stay in orbit.

Q: What is the total energy of a satellite in a circular orbit?

For any circular orbit, the total mechanical energy is exactly half of its potential energy, given by $E = -GMm / 2r$.

Q: Is the launch energy the same as the escape energy?

No. Escape energy is the energy required to send the satellite to infinity where its final total energy is zero. Here, the satellite remains bound to the planet in an orbit, so its final total energy is negative.

Question

What is the minimum energy required to launch a satellite of mass $m$ from the surface of a planet of mass $M$ and radius $R$ in a circular orbit at an altitude of $2R$?

Options:

(A) $\frac{5GmM}{6R}$

(B) $\frac{2GmM}{3R}$

(C) $\frac{GmM}{2R}$

(D) $\frac{GmM}{3R}$

Sigma Percentile · Accepted Answer

Launching a satellite into orbit is one of the most classic applications of the conservation of mechanical energy. Let's break down the physics behind this maneuver step-by-step.

### Analyzing the Setup

Imagine a planet of mass $M$ and radius $R$. We have a satellite of mass $m$ resting on its surface. Our goal is to launch it into a stable circular orbit at an altitude of $2R$.

It is crucial to remember that gravitational potential energy and orbital mechanics depend on the distance from the **center** of the planet. Therefore, the total orbital radius $r$ is the sum of the planet's radius and the altitude:

$$r = R + 2R = 3R$$

### The Master Equation

To find the minimum launch energy, we rely on the principle of conservation of mechanical energy. The total energy of the satellite on the surface, plus the kinetic energy we provide to launch it ($K_{	ext{launch}}$), must equal its total mechanical energy in the final orbit.

$$E_{	ext{surface}} + K_{	ext{launch}} = E_{	ext{orbit}}$$

On the surface, the satellite is at rest, so it only possesses gravitational potential energy:

$$E_{	ext{surface}} = -\frac{GMm}{R}$$

In the orbit, the satellite possesses both potential energy and kinetic energy (since it must travel at a specific orbital speed $v_0$ to avoid falling back):

$$E_{	ext{orbit}} = -\frac{GMm}{3R} + \frac{1}{2}mv_0^2$$

### Finding the Orbital Energy

To find the orbital kinetic energy, we use the condition for a stable circular orbit: the gravitational force provides the necessary centripetal force.

$$\frac{mv_0^2}{3R} = \frac{GMm}{(3R)^2}$$

Solving for $v_0^2$, we get:

$$v_0^2 = \frac{GM}{3R}$$

Substituting this back into our kinetic energy expression:

$$K_{	ext{orbit}} = \frac{1}{2}m\left(\frac{GM}{3R}ight) = \frac{GMm}{6R}$$

Now, we can find the total energy in the orbit by adding the potential and kinetic energies:

$$E_{	ext{orbit}} = -\frac{GMm}{3R} + \frac{GMm}{6R} = -\frac{2GMm}{6R} + \frac{GMm}{6R} = -\frac{GMm}{6R}$$

*(Pro Tip: For any circular orbit, the total mechanical energy is always exactly half of its potential energy, i.e., $E = -\frac{GMm}{2r}$. Since $r = 3R$, $E = -\frac{GMm}{6R}$. Knowing this shortcut saves precious time!)*

### Final Calculation

Finally, the launch energy required is the difference between the final total energy and the initial potential energy:

$$K_{	ext{launch}} = E_{	ext{orbit}} - E_{	ext{surface}}$$

$$K_{	ext{launch}} = -\frac{GMm}{6R} - \left(-\frac{GMm}{R}ight)$$

$$K_{	ext{launch}} = \frac{GMm}{R} - \frac{GMm}{6R} = \frac{6GmM}{6R} - \frac{GmM}{6R} = \frac{5GmM}{6R}$$

This is the minimum kinetic energy the rocket launcher must provide to successfully place the satellite into its designated orbit.

Animated Solution for Physics - Gravitation: What is the minimum energy required to launch a satellite of mass $m$ from the surface of a planet of mass $M$ and radius $R$ in a circular orbit at an altitude of $2 R$ ?

Select Answer:

Visualized Solution

$r_{orbit} = R + 2 R = 3 R$

$E_{surface} + K_{launch} = E_{orbit}$

$U_{surface} + K_{launch} = U_{orbit} + K_{orbit}$

$\frac{m v _{0}^{2}}{3 R} = \frac{GM m}{( 3 R ) ^{2}}$

$E_{orbit} = - \frac{GM m}{6 R}$

$K_{launch} = \frac{5 G m M}{6 R}$

$E_{orbit} = - \frac{GM m}{2 r}$

The Sigma Insight: Orbital Motion of a Satellite

Analyzing the Setup

The Master Equation

Finding the Orbital Energy

Final Calculation

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