Gravitational potential is the work done per unit mass when a mass is brought from infinity to a point in space at a small constant speed.

As force is not constant, the work done is found by integrating the force with respect to the distance from infinity.

\(V_g= {\int_\infty^r F_g\,\mathrm{d}r\over m}={\int_\infty^r {GMm\over r^2}\,\mathrm{d}r\over m}\)

\(\Rightarrow V_g=-G{M\over r}\)

• \(V_g\) is gravitational potential (Nkg^-1)
• \(G\) is the universal gravitational constant (6.67 × 10^-11 m³kg^-1s^-2)
• \(M\) is the mass producing the field (kg)
• \(r\) is the distance between the centers of the masses (m)

NB: Gravitational potential is negative as energy is released as the mass is brought from infinity to within the field. This is because gravitational forces are attractive.

Since gravitational potential varies with \(1\over r\) from the center of the mass, the equipotential surfaces for a radial field increase in spacing outward from the mass producing the field. They are always at right angles to the field lines (shown in red):

No energy is changed when the mass moves along an equipotential surface.

When a mass moves parallel to a field line, its potential energy changes. Potential energy is the product of potential and the mass placed in the field:

\(E_p=mV_g=-G{Mm\over r}\)

Change in potential energy can be calculated as follows:

\(\Delta E_p=m\Delta V_g=m\times GM({1\over r_f}-{1\over r_i})\)

• \(\Delta E_p\) is change in potential energy (J)
• \(m\) is the mass moving in the field (kg)
• \(\Delta V_g\) is the change in potential (Jkg^-1)
• \(G\) is the universal gravitational constant (6.67 × 10^-11 m³kg^-1s^-2)
• \(M\) is the mass producing the field (kg)
• \(r\) is the initial or final distance between the centers of the masses (m)

Potential gradient between two equipotential surfaces is the ratio of the potential difference to the distance between the equipotentials:

\({\Delta V_g\over \Delta r}=-G{M\over r^2}\)

Notice that potential gradient is equal to negative gravitational field strength:

\(g=-{\Delta V_g\over \Delta r} =G{M\over r^2}\)

• \(g\) is gravitational field strength (Nkg^-1)
• \(\Delta V_g\) is the change in potential (Jkg^-1)
• \(\Delta r\) is the distance between the equipotentials (m)
• \(G\) is the universal gravitational constant (6.67 × 10^-11 m³kg^-1s^-2)
• \(M\) is the mass producing the field (kg)
• \(r\) is the distance between the centers of masses (m)

A mass in orbit in a gravitational field possesses two types of energy:

• Kinetic energy due to its motion, \(E_k={1\over 2}mv^2\)
• Gravitational potential energy due to its position in the field, \(E_p=-G{Mm\over r}\)

Total energy, \(E_T=E_k+E_p\):

\(E_T={1\over 2}mv^2-G{Mm\over r}\)

We can reduce the number of variables by considering the implication of circular orbital motion, that the resultant centripetal force is equal to the gravitational force:

\(F_g=G{Mm\over r^2}=m{v^2\over r}\)

\(v^2=G{M\over r}\)

Substituting into the total energy:

\(E_T={1\over 2}G{Mm\over r}-G{Mm\over r}=-{1\over 2}G{Mm\over r}\)

NB: We bypassed a Data Booklet equation en route, which gives the size of the velocity for an orbtting body:

\(v_\text{orbit}=\sqrt{G{M\over r}}\)

A ballistic object (i.e. a particle without its own engine) can theoretically be given sufficient kinetic energy to fully escape the gravitational field of the mass on which it is situated. The escape speed is the speed that would just be enough to escape the field.

To derive an expression for this escape velocity, we use conservation of energy. The kinetic energy at the surface added to the gravitational potential energy at the surface is equal to the energy at infinity. The energy at infinity is zero because the object will just have lost its kinetic energy and because of how gravitational potential is defined when the field is escaped.

\(E_k+E_p=0\)

\({1\over 2}mv^2-G{Mm\over r}=0\)

\({1\over 2}mv^2=G{Mm\over r}\)

\(\Rightarrow v_\text{esc}=\sqrt{2G{M\over r}}\)