Conservation of Energy
Conservation of Energy
In Potential Energy and Conservation of Energy, we described how to apply conservation of energy for systems with conservative forces. We were able to solve many problems, particularly those involving gravity, more simply using conservation of energy. Those principles and problem-solving strategies apply equally well here. The only change is to place the new expression for potential energy into the conservation of energy equation, \(E={K}_{1}+{U}_{1}={K}_{2}+{U}_{2}\).
Note that we use M, rather than \({M}_{\text{E}}\), as a reminder that we are not restricted to problems involving Earth. However, we still assume that \(m\text{<}\;\text{<}M\). (For problems in which this is not true, we need to include the kinetic energy of both masses and use conservation of momentum to relate the velocities to each other. But the principle remains the same.)
Escape velocity
Escape velocity is often defined to be the minimum initial velocity of an object that is required to escape the surface of a planet (or any large body like a moon) and never return. As usual, we assume no energy lost to an atmosphere, should there be any.
Consider the case where an object is launched from the surface of a planet with an initial velocity directed away from the planet. With the minimum velocity needed to escape, the object would just come to rest infinitely far away, that is, the object gives up the last of its kinetic energy just as it reaches infinity, where the force of gravity becomes zero. Since \(U\to 0\;\text{as}\;r\to \infty \), this means the total energy is zero. Thus, we find the escape velocity from the surface of an astronomical body of mass M and radius R by setting the total energy equal to zero. At the surface of the body, the object is located at \({r}_{1}=R\) and it has escape velocity \({v}_{1}={v}_{\text{esc}}\). It reaches \({r}_{2}=\infty \) with velocity \({v}_{2}=0\). Substituting into Equation, we have
\(\cfrac{1}{2}m{v}_{\text{esc}}^{2}-\cfrac{GMm}{R}=\cfrac{1}{2}m{0}^{2}-\cfrac{GMm}{\infty }=0.\)
Solving for the escape velocity,
Notice that m has canceled out of the equation. The escape velocity is the same for all objects, regardless of mass. Also, we are not restricted to the surface of the planet; R can be any starting point beyond the surface of the planet.
Example: Escape from Earth
What is the escape speed from the surface of Earth? Assume there is no energy loss from air resistance. Compare this to the escape speed from the Sun, starting from Earth’s orbit.
Strategy
We use Equation, clearly defining the values of R and M. To escape Earth, we need the mass and radius of Earth. For escaping the Sun, we need the mass of the Sun, and the orbital distance between Earth and the Sun.
Solution
Substituting the values for Earth’s mass and radius directly into Equation, we obtain
\({v}_{\text{esc}}=\sqrt{\cfrac{2GM}{R}}=\sqrt{\cfrac{2(6.67\;×\;{10}^{-11}\;\text{N}·{\text{m}}^{2}{\text{/kg}}^{2})(5.96\;×\;{10}^{24}\;\text{kg})}{6.37\;×\;{10}^{6}\;\text{m}}}=1.12\;×\;{10}^{4}\;\text{m/s.}\)
That is about 11 km/s or 25,000 mph. To escape the Sun, starting from Earth’s orbit, we use \(R={R}_{\text{ES}}=1.50\;×\;{10}^{11}\;\text{m}\) and \({M}_{\text{Sun}}=1.99\;×\;{10}^{30}\;\text{kg}\). The result is \({v}_{\text{esc}}=4.21\;×\;{10}^{4}\;\text{m/s}\) or about 42 km/s.
Significance
The speed needed to escape the Sun (leave the solar system) is nearly four times the escape speed from Earth’s surface. But there is help in both cases. Earth is rotating, at a speed of nearly 1.7 km/s at the equator, and we can use that velocity to help escape, or to achieve orbit. For this reason, many commercial space companies maintain launch facilities near the equator. To escape the Sun, there is even more help. Earth revolves about the Sun at a speed of approximately 30 km/s. By launching in the direction that Earth is moving, we need only an additional 12 km/s. The use of gravitational assist from other planets, essentially a gravity slingshot technique, allows space probes to reach even greater speeds. In this slingshot technique, the vehicle approaches the planet and is accelerated by the planet’s gravitational attraction. It has its greatest speed at the closest point of approach, although it decelerates in equal measure as it moves away. But relative to the planet, the vehicle’s speed far before the approach, and long after, are the same. If the directions are chosen correctly, that can result in a significant increase (or decrease if needed) in the vehicle’s speed relative to the rest of the solar system.
Visit this website to learn more about escape velocity of different planets in our solar system.
This lesson is part of:
Uniform Circular Motion and Gravitation