Escape Velocity: Videos & Practice Problems

Escape velocity is the minimum speed required for an object to break free from a gravitational field without returning. It is derived from the conservation of energy, where the initial kinetic energy equals the gravitational potential energy at a distance where gravity approaches zero. The formula for escape velocity is $\sqrt{\frac{2}{gM}r}$ , depending only on the mass of the celestial body and the distance from its center. For example, to escape the Sun from Earth's distance, the required speed is approximately 42.1 km/s.

Escape Velocity

Escape velocity is the minimum speed required for an object to break free from the gravitational pull of a celestial body, such as Earth or the Sun, without any further propulsion. This concept is crucial in understanding how objects can leave a planet's gravitational influence and travel into space. When an object reaches escape velocity, it will continue moving away from the planet until the gravitational force approaches zero, meaning it will not return.

The formula for escape velocity (\(v_{\text{escape}}\)) can be derived from the conservation of energy principle, which states that the total mechanical energy (kinetic plus potential) remains constant in a closed system. The equation can be expressed as:

\[ v_{\text{escape}} = \sqrt{\frac{2GM}{r}} \]

In this equation, \(G\) represents the gravitational constant (\(6.67 \times 10^{-11} \, \text{m}^3/\text{kg} \cdot \text{s}^2\)), \(M\) is the mass of the celestial body being escaped from, and \(r\) is the distance from the center of that body to the point of launch.

When an object is thrown upwards, it initially possesses kinetic energy, given by \(KE = \frac{1}{2}mv^2\), and gravitational potential energy, which is negative and given by \(PE = -\frac{GMm}{r}\). As the object ascends, it loses kinetic energy and gains potential energy until it reaches a point where its final velocity is zero, indicating that it has escaped the gravitational influence of the body.

Importantly, the escape velocity is independent of the mass of the object attempting to escape; it solely depends on the mass of the celestial body and the distance from its center. For example, to calculate the escape velocity for a 5-kilogram rock thrown from Earth's orbital distance to escape the Sun, one would use the average distance from the Earth to the Sun (\(1.5 \times 10^{11} \, \text{m}\)) and the mass of the Sun (\(2 \times 10^{30} \, \text{kg}\)). Plugging these values into the escape velocity formula yields:

\[ v_{\text{escape}} = \sqrt{2 \cdot (6.67 \times 10^{-11}) \cdot (2 \times 10^{30}) / (1.5 \times 10^{11})} \approx 42.1 \, \text{km/s} \]

This means that to escape the Sun's gravitational pull from Earth's distance, the rock must be thrown at a speed of approximately 42.1 kilometers per second. Understanding escape velocity is essential for space missions and satellite launches, as it determines the energy required to leave a planet's gravitational field.

a) How fast does a spaceship have to go to escape Earth, starting from the launch pad on the surface? Assume it burns all its fuel very fast and then shuts off the engines.
b) One idea to make getting to space easier is to build a space elevator, a large platform high above Earth's surface where spaceships can land and take off. How high above Earth would this platform have to be for the escape velocity to be 1/5 of its surface value? (For multiple choice, select the correct answer for part (b).)

• A. $2.52\times 10^3$ m
• B. $1.53\times 10^8$ m
• C. $1.59\times 10^8$ m
• D. $7.11\times 10^{10}$ m

Answer: (B)

Jumping Off an Asteroid

When considering the escape velocity from a large spherical asteroid, we utilize the formula for escape velocity, which is given by:

\( v_{\text{escape}} = \sqrt{\frac{2GM}{r}} \)

In this equation, \( G \) represents the gravitational constant, approximately \( 6.67 \times 10^{-11} \, \text{m}^3/\text{kg} \cdot \text{s}^2 \), \( M \) is the mass of the asteroid, and \( r \) is the radius from the center of the asteroid to the point of escape.

To find the escape velocity, we first need to determine the radius of the asteroid. If you walk around the asteroid and measure a distance of 25 kilometers (or 25,000 meters), this distance corresponds to the circumference of the asteroid. The relationship between circumference \( C \) and radius \( r \) is given by:

\( C = 2\pi r \)

From this, we can rearrange to find the radius:

\( r = \frac{C}{2\pi} = \frac{25000}{2\pi} \approx 3980 \, \text{meters} \)

Next, we need to find the mass of the asteroid. To do this, we can analyze the motion of an object dropped from a height of 1.4 meters, which takes 30 seconds to hit the ground. Using the kinematic equation for vertical motion:

\( \Delta y = v_{\text{initial}} t + \frac{1}{2} g t^2 \)

Since the initial velocity is zero, this simplifies to:

\( \Delta y = \frac{1}{2} g t^2 \)

Substituting the known values:

\( 1.4 = \frac{1}{2} g (30^2) \)

Rearranging gives us:

\( g = \frac{2 \times 1.4}{30^2} \approx 0.00311 \, \text{m/s}^2 \)

Now, we can relate this gravitational acceleration to the mass of the asteroid using the formula:

\( g = \frac{GM}{r^2} \)

Rearranging for mass \( M \) gives:

\( M = \frac{g r^2}{G} \)

Substituting the values we have:

\( M = \frac{0.00311 \times (3980)^2}{6.67 \times 10^{-11}} \approx 7.39 \times 10^{14} \, \text{kg} \)

With both the mass and radius determined, we can now calculate the escape velocity:

\( v_{\text{escape}} = \sqrt{\frac{2 \times (6.67 \times 10^{-11}) \times (7.39 \times 10^{14})}{3980}} \approx 4.98 \, \text{m/s} \)

This result indicates that to escape the asteroid's gravitational pull, one would need to jump at a speed of approximately 4.98 meters per second, a feasible task even in a spacesuit.

Here’s what students ask on this topic:

What is escape velocity and how is it calculated?

Escape velocity is the minimum speed required for an object to break free from a gravitational field without returning. It is derived using the conservation of energy principle, where the initial kinetic energy equals the gravitational potential energy at a distance where gravity approaches zero. The formula for escape velocity is:

$\sqrt{\frac{2}{𝛂𝛱}}$

where $𝛂$ is the gravitational constant, $𝛱$ is the mass of the celestial body, and $𝑛$ is the distance from the center of the celestial body.

Why does escape velocity not depend on the mass of the object trying to escape?

Escape velocity does not depend on the mass of the object trying to escape because the mass cancels out during the derivation of the escape velocity formula. The formula is derived from the conservation of energy, where the initial kinetic energy and gravitational potential energy are set equal. Since both kinetic energy ($\frac{1}{2}𝑚{𝑛}^2$) and gravitational potential energy ($-𝛂𝛱𝑚𝑛∕𝑛$) include the mass of the object, it cancels out, leaving the escape velocity dependent only on the mass of the celestial body and the distance from its center.

How fast must an object be thrown to escape Earth's gravity?

To escape Earth's gravity, an object must be thrown with a speed of approximately 11.2 km/s. This speed is derived using the escape velocity formula:

$\sqrt{\frac{2}{𝛂𝛱}}$

where $𝛂$ is the gravitational constant (6.67 × 10⁻¹¹ N(m/kg)²), $𝛱$ is the mass of Earth (5.97 × 10²⁴ kg), and $𝑛$ is the radius of Earth (6.37 × 10⁶ m).

What is the escape velocity from the Sun at Earth's orbital distance?

The escape velocity from the Sun at Earth's orbital distance is approximately 42.1 km/s. This is calculated using the escape velocity formula:

$\sqrt{\frac{2}{𝛂𝛱}}$

where $𝛂$ is the gravitational constant (6.67 × 10⁻¹¹ N(m/kg)²), $𝛱$ is the mass of the Sun (2 × 10³⁰ kg), and $𝑛$ is the average distance from the Earth to the Sun (1.5 × 10¹¹ m).

How does the escape velocity change with distance from the celestial body?

Escape velocity decreases as the distance from the celestial body increases. This relationship is described by the escape velocity formula:

$\sqrt{\frac{2}{𝛂𝛱}}$

where $𝑛$ is the distance from the center of the celestial body. As $𝑛$ increases, the value under the square root decreases, leading to a lower escape velocity. This is because gravitational force weakens with distance, requiring less speed to overcome it.