The gravitational potential energy Ep is given by the following formula:
$$\begin{aligned} \mathrm{E_p = - \frac {Gm_1m_2}{r}} \end{aligned}$$
How does this formula compare to Ep= mgh ?
In more elementary physics courses we used a more simple formula, which was Ep=mgh. This formula is valid when the height h is not very large, e.g., the mass is relatively close to the surface of the planet. It can be demonstrated mathematically that both formula for gravitational potential energy are approximately equal in the limit for relatively low heights.
Why is potential energy negative?
The idea is that when the distance (r) is large enough, there will be no interaction between the masses 1 and 2 so that their potential energy is zero. If the value is less than zero the particles are attracting each other. For instance, it could be stars forming a binary system or it could be atoms forming a molecule.
This formula is very useful to solve problems using the principle of conservation of energy. The total mechanical energy is the sum of the kinetic and potential energies.
Kinetic energy is given , as usual, by:
$$\begin{aligned} \mathrm{E_k = \frac {mv^2}{2}} \end{aligned}$$
So that the total mechanical energy of the object is given by:
$$\begin{aligned} \mathrm{E_T = E_k+ E_p } \end{aligned}$$
or
$$\begin{aligned} \mathrm{E_{T1} = \frac {m_1v_1^2}{2} - \frac {Gm_1m_2}{r}} \end{aligned}$$
and similarly,
$$\begin{aligned} \mathrm{E_{T2} = \frac {m_2v_2^2}{2} - \frac {Gm_1m_2}{r}} \end{aligned}$$
What if the total energy is negative?
That means we have a bound state, e.g., the particles are pulling each other with enough strength to be kept together. It could be planet orbiting a star or electrons moving around a central nucleus. If the kinetic energy of the planet is high enough it can leave the orbit and go away in a straight line. If its kinetic energy is not high enough to overcome the potential energy (of gravitational attraction) it stays bound.
What is the difference between gravitational potential energy and gravitational potential?
Gravitational potential is the gravitational potential energy divides by the mass. This formula appears on the IB data booklet on sub-topic 10.2-Fields at work. It is represented by Vg :
$$\begin{aligned} \mathrm{V_g = - \frac {Gm}{r}} \end{aligned}$$
Example calculations
Example 1 - The total energy of the Earth orbit
Data:
average speed: 29.78 * 103m/s
average distance from the Sun: 1.50 * 1011 m
mass: 5.98 * 1024 kg
According to data provided by NASA and and also calculations displayed on the spreadsheet below, we have for the Earth:
Ek= 2650.7994 * 1030 J
Ep= -5294.236705 * 1030 J
ET= 2988.334621 * 1030 J
As expected , the total energy is negative so that the Earth is bound to the Sun!
Theses calculations are also performed for all planets . Calculations of escape velocity are also performed, using the formula that is deduced on the next example.
To check the formulas used the worksheet can be accessed here (Google sheets)
Example 2 - The calculation of escape velocity
Escape velocity is the minimum velocity needed to escape a planet (or anything). For instance, if you want your rocket to get out of the Earth, you need to take off and travel at escape velocity. Because this is the minimum velocity to escape, we assume that the velocity after escape is zero, so that Ek is also zero. Ep must also be zero, because the rocket will be free from the gravitational attraction of the Earth. The EK was transferred into Ep. So Ek must equal Ep at the surface of the planet.
Another strategy is to use the PCE (principle of conservation of energy). The total energy before the launch and after leaving the planet must be the same. The total energy after leaving the planet is zero, as explained above.
So let´s calculate the escape velocity of the Earth:
$$\begin{aligned} \mathrm{0 = \frac {m_2v_2^2}{2} - \frac {Gm_1m_2}{r}} \end{aligned}$$
m2 = mass of the rocket
m1= mass of the Earth
r = radius of the Earth
$$\begin{aligned} \mathrm{\frac {m_2v_2^2}{2} = \frac {Gm_1m_2}{r}} \end{aligned}$$
m2 is on both sides, so that it can be cancelled. It is interesting that it doesn´t depend on the mass of the rocket or whatever object is being ejected from the planet. Finally:
$$\begin{aligned} \mathrm{v_{escape} = \sqrt\frac {2Gm_1}{r}} \end{aligned}$$
This formula works for any planet. I have calculated it for all the planets in the solar system (spreadsheet above)