Calculating Escape Velocity
The escape velocity vesc is expressed as vesc = Square root of√2GMr,where G is the gravitational constant, M is the mass of the attracting mass, and r is the distance from the centre of that mass.
Escape velocity equation
The escape velocity formula is independent of the properties of the escaping object. The only thing that matters is the mass and radius of the celestial body in question:
The escape velocity formula is independent of the properties of the escaping object. The only thing that matters is the mass and radius of the celestial body in question:
\[ V_e = \sqrt{\frac{2GM}{R}} \]
Where:
- M is the mass of the planet,
- R is its radius, and
- G is the gravitational constant, which is equal to \( G = 6.674 \times 10^{-11} \, \text{N} \cdot \text{m}^2/\text{kg}^2 \).
The formula for escape velocity is derived directly from the law of conservation of energy. At the moment of launch, the object has some potential energy \( PE \) and some kinetic energy \( KE \). The total energy at launch \( LE \) can be presented as follows:
\[ PE + KE = -\frac{GMm}{R} + \frac{1}{2}mv^2 \]
Where:
- m is the mass of the object being launched,
- v is the escape velocity.
When the object finally escapes, it is so far from the planet that its potential energy is zero. Its kinetic energy is also zero, meaning the total energy is zero:
\[ PE + KE = 0 + 0 = 0 \]
Since the total energy must be conserved, it means that the initial energy is also equal to zero. Simplifying the first equation, we get:
\[ 0 = -\frac{GMm}{R} + \frac{1}{2}mv^2 \]
Solving for \( v \), we get the escape velocity formula:
\[ v = \sqrt{\frac{2GM}{R}} \]
Example
Calculating Escape Velocity
The escape velocity refers to the minimum speed an object must have in order to break free from the gravitational pull of a celestial body, without further propulsion. It depends on the mass and radius of the body and is independent of the object's properties.
The general approach to calculating escape velocity includes:
- Identifying the celestial body from which the object is escaping.
- Knowing the mass and radius of the body.
- Applying the escape velocity formula to calculate the minimum speed required.
Escape Velocity Formula
The general formula for escape velocity is:
\[ V_e = \sqrt{\frac{2GM}{R}} \]
Where:
- Ve is the escape velocity (in m/s).
- G is the gravitational constant, \( G = 6.674 \times 10^{-11} \, \text{N} \cdot \text{m}^2/\text{kg}^2 \).
- M is the mass of the celestial body (in kg).
- R is the radius of the celestial body (in meters).
Example:
If a rocket needs to escape from a planet with a mass of \( 5 \times 10^{24} \, \text{kg} \) and a radius of \( 6.4 \times 10^6 \, \text{m} \), the escape velocity is:
- Step 1: Use the escape velocity formula: \( V_e = \sqrt{\frac{{(6.674 \times 10^{-11})(5 \times 10^{24})}}{{6.4 \times 10^6}}} \).
- Step 2: Calculate the result: \( V_e = \sqrt{\frac{{(6.674 \times 10^{-11})(5 \times 10^{24})}}{{6.4 \times 10^6}}} = 11,200 \, \text{m/s} \).
Escape Velocity for Different Celestial Bodies
The escape velocity varies based on the mass and radius of the celestial body. For example, the escape velocity from Earth is about 11.2 km/s, while from the Moon, it is about 2.4 km/s.
Example:
If the mass of the Earth is \( 5.97 \times 10^{24} \, \text{kg} \) and its radius is \( 6.38 \times 10^6 \, \text{m} \), the escape velocity is approximately 11.2 km/s, which can be calculated using the escape velocity formula.
Real-life Applications of Escape Velocity
Escape velocity plays a crucial role in various fields such as:
- Space exploration and the launch of rockets and satellites.
- Understanding the movement of celestial objects, such as comets and asteroids.
- Designing space missions to escape Earth's gravitational pull and reach other planets or the moon.
Common Units for Escape Velocity
SI Unit: The unit of escape velocity is meters per second (m/s).
Escape velocity depends on the mass and radius of the celestial body and is independent of the mass of the object attempting to escape.
Common Operations with Escape Velocity
Escape from Different Celestial Bodies: The escape velocity changes based on the mass and radius of the celestial body.
Escape in a Vacuum: Escape velocity assumes there is no resistance from the atmosphere, as the object travels through the vacuum of space.
Escape with Thrust: The formula for escape velocity assumes no additional propulsion, though in real missions, rockets must continue accelerating to reach the required speed.
| Problem Type | Description | Steps to Solve | Example |
|---|---|---|---|
| Escape Velocity for a Planet | Finding the escape velocity required to leave a planet's gravitational field. |
|
If the mass of the planet is \( 5 \times 10^{24} \, \text{kg} \) and its radius is \( 6.4 \times 10^6 \, \text{m} \), the escape velocity is \( V_e = \sqrt{\frac{2(6.674 \times 10^{-11})(5 \times 10^{24})}{6.4 \times 10^6}} = 11,200 \, \text{m/s} \). |
| Escape Velocity for a Moon | Finding the escape velocity from the gravitational field of a moon. |
|
If the mass of the moon is \( 7.35 \times 10^{22} \, \text{kg} \) and its radius is \( 1.74 \times 10^6 \, \text{m} \), the escape velocity is \( V_e = \sqrt{\frac{2(6.674 \times 10^{-11})(7.35 \times 10^{22})}{1.74 \times 10^6}} = 2,420 \, \text{m/s} \). |
| Escape Velocity for a Star | Finding the escape velocity required to escape the gravitational field of a star. |
|
If the mass of the star is \( 1.989 \times 10^{30} \, \text{kg} \) and its radius is \( 6.96 \times 10^8 \, \text{m} \), the escape velocity is \( V_e = \sqrt{\frac{2(6.674 \times 10^{-11})(1.989 \times 10^{30})}{6.96 \times 10^8}} = 617,000 \, \text{m/s} \). |
| Escape Velocity for a Black Hole | Finding the escape velocity at the event horizon of a black hole. |
|
If the mass of the black hole is \( 10^6 \, \text{M}_\odot \) (solar masses) and the Schwarzschild radius is calculated using \( R_s = \frac{2GM}{c^2} \), the escape velocity at the event horizon is equal to the speed of light \( V_e = c \), approximately \( 3 \times 10^8 \, \text{m/s} \). |
