# Escape Speed

Show/Hide Sub-topics (Gravitation | A Level Physics)

## Escape Speed

Escape speed, also known as escape velocity, is the minimum speed an object must reach to break free from the gravitational pull of a celestial body (or typically in local context, Earth’s), without further propulsion.

\begin{aligned} V_{\text{escape}} \, &= \, \sqrt{2g R_{E}} \\ & OR \\ V_{\text{escape}} &= \sqrt{\frac{2GM}{R}} \end{aligned}

, where

• $V$ is escape speed,
• $g$ is gravitational field strength,
• $R$ is radius of the Earth,
• $M$ is the mass of Earth,
• $R$ is the radius of the Earth,
• $G$ is the gravitational constant ($6.674 \times 10^{-11} \text{ Nm}^{2}\text{kg}^{-2}$).

## Derivation of Escape Speed From Earth

We know that:

$$\text{Total Energy at infinity} = 0$$

Hence,

\begin{aligned} \text{Kinetic energy } + \text{ Potential energy} &= 0 \\ \frac{1}{2} m v^{2} + \left( – \frac{\left( G M m \right)}{R} \right) &=0 \\ \frac{1}{2} m v^{2} &= \frac{ G M m}{R} \\ v^{2} &= 2 \frac{G M}{R} \end{aligned}

,where

• $v$ is the velocity of the object,
• $m$ is the mass of the object,
• $M$ is the mass of Earth,
• $R$ is the radius of the Earth,
• $G$ is the gravitational constant ($6.674 \times 10^{-11} \text{ Nm}^{2}\text{kg}^{-2}$).

From Gravitational Field Strength, we know that $g = G \frac{M}{R^{2}}$. Substitute this into the equation above, we will have:

$$v^{2} = 2 gR$$

In the context of this derivation, we have:

$$V_{\text{escape}} \, = \, \sqrt{2g R_{E}}$$

## Implications of Escape Speed

### Space Exploration

The concept of escape speed is fundamental to launching spacecraft from Earth or any other celestial body. For instance, Earth’s escape speed is approximately 11.2 kilometers per second (km/s). This means any spacecraft must reach or exceed this speed to break free from Earth’s gravitational influence and venture into deep space.

### Cosmic Perspective

Escape speed also offers insights into celestial dynamics, such as the conditions necessary for a planet to retain an atmosphere. Bodies with escape speeds significantly lower than the thermal speeds of atmospheric molecules tend to lose those molecules into space, explaining why smaller celestial bodies, like the Moon, have thin or no atmospheres.

### Black Holes

The concept reaches its extreme in the form of black holes, where the escape speed exceeds the speed of light. This means not even light can escape the gravitational pull of a black hole, rendering these cosmic entities invisible and highlighting the profound effects of gravity.

## Worked Examples

### Example 1: Earth’s Escape Speed

A spacecraft aims to leave Earth’s gravitational pull from the surface. Given that Earth’s mass is approximately $5.972 \times 10^{24} \text{ kg}$ and its radius is about 6,371 km, calculate the escape speed from Earth.

To find the escape speed, we use the formula:
$$v_{esc} = \sqrt{\frac{2G M}{R}}$$

Where:

• $G = 6.674 \times 10^{-11} \, \text{m}^3\text{kg}^{-1}\text{s}^{-2}$ (universal gravitational constant),
• $M = 5.972 \times 10^{24} \, \text{kg}$ (mass of Earth),
• $R = 6,371 \, \text{km} = 6,371,000 \, \text{m}$ (radius of Earth).

Substituting these values into the formula gives:

\begin{aligned} v_{esc} &= \sqrt{\frac{2 \times 6.674 \times 10^{-11} \times 5.972 \times 10^{24}}{6,371,000}} \\ &= \sqrt{\frac{7.977 \times 10^{14}}{6,371,000}} \\ &= \sqrt{125.1 \times 10^{6}} \\ &\approx 11.2 \, \text{km/s} \end{aligned}

Therefore, the escape speed from Earth is approximately 11.2 km/s.

### Example 2: Moon’s Escape Speed

Question:
The Moon’s mass is about $7.347 \times 10^{22}$ kg, and its radius is roughly 1,737 km. Calculate the escape speed from the Moon’s surface.

Using the escape speed formula:

$$v_{esc} = \sqrt{\frac{2G M}{R}}$$

Where:

• $M = 7.347 \times 10^{22} \, \text{kg}$,
• $R = 1,737 \, \text{km} = 1,737,000 \, \text{m}$.

Substituting these values gives:

\begin{aligned} v_{esc} &= \sqrt{\frac{2 \times 6.674 \times 10^{-11} \times 7.347 \times 10^{22}}{1,737,000}} \\ &= \sqrt{\frac{9.81 \times 10^{12}}{1,737,000}} \\ &= \sqrt{5.65 \times 10^{6}} \\ &\approx 2.38 \, \text{km/s} \end{aligned}

Thus, the escape speed from the Moon is approximately 2.38 km/s.

### Example 3: Jupiter’s Escape Speed

Question:
Jupiter is the largest planet in our solar system, with a mass of $1.898 \times 10^{27} \text{ kg}$ and a radius of 69,911 km. What is the escape speed from Jupiter’s surface?

Applying the formula:

$$v_{esc} = \sqrt{\frac{2G M}{R}}$$

Where:

• $M = 1.898 \times 10^{27} \, \text{kg}$,
• $R = 69,911 \, \text{km} = 69,911,000 \, \text{m}$.

Substituting these values gives:

\begin{aligned} v_{esc} &= \sqrt{\frac{2 \times 6.674 \times 10^{-11} \times 1.898 \times 10^{27}}{69,911,000}} \\ &= \sqrt{\frac{2.535 \times 10^{17}}{69,911,000}} \\ &= \sqrt{3.626 \times 10^{9}} \\ & \approx 60.2 \, \text{km/s} \end{aligned}

Therefore, the escape speed from Jupiter is approximately 60.2 km/s.

### Example 4: A Theoretical Planet

A newly discovered exoplanet has twice the mass of Earth and half its radius. What is the escape speed from the surface of this exoplanet?

Let’s denote Earth’s mass by $M_{\oplus}$ and its radius by $R_{\oplus}$. The mass of the exoplanet is $2M_{\oplus}$ and its radius is $0.5R_{\oplus}$.

The escape speed formula is:

$$v_{esc} = \sqrt{\frac{2G M}{R}}$$

Substituting the exoplanet’s mass and radius gives:

\begin{aligned} v_{esc} &= \sqrt{\frac{2G (2M_{\oplus})}{0.5R_{\oplus}}} \\ &= \sqrt{\frac{4G M_{\oplus}}{0.5R_{\oplus}}} \\ &= \sqrt{\frac{8G M_{\oplus}}{R_{\oplus}}} \end{aligned}

Since the escape speed from Earth is known to be approximately 11.2 km/s, we have:

$$11.2 = \sqrt{\frac{2G M_{\oplus}}{R_{\oplus}}}$$

Therefore, the escape speed from the exoplanet would be:

\begin{aligned} v_{esc} &= 11.2 \times \sqrt{4} \\ &= 11.2 \times 2 \\ &= 22.4 \, \text{km/s} \end{aligned}

The escape speed from this theoretical exoplanet is 22.4 km/s.

### Example 5: Satellite Escape Speed

A satellite orbits a planet similar to Earth at an altitude of 300 km above the surface. What is the escape speed for this satellite?

The total distance from the planet’s center is the sum of Earth’s radius and the satellite’s altitude:

\begin{aligned} R_{total} &= R_{\oplus} + 300 \, \text{km} \\ &= 6,371 \, \text{km} + 300 \, \text{km} \\ &= 6,671 \, \text{km} = 6,671,000 \, \text{m} \end{aligned}

Using the escape speed formula:

$$v_{esc} = \sqrt{\frac{2G M}{R_{total}}}$$

Substituting Earth’s mass and the total radius gives:

\begin{aligned} v_{esc} &= \sqrt{\frac{2 \times 6.674 \times 10^{-11} \times 5.972 \times 10^{24}}{6,671,000}} \\ &= \sqrt{\frac{7.977 \times 10^{14}}{6,671,000}} \\ &= \sqrt{119.6 \times 10^{6}} \\ & \approx 10.95 \, \text{km/s} \end{aligned}

Hence, the escape speed for the satellite at an altitude of 300 km above Earth’s surface is approximately 10.95 km/s.

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