Gravitational Potential Energy
- The gravitational potential Vg at a point is given by:
- Recall that potential V is defined as the energy required to bring unit mass m from infinity to a defined point in the field
- Recall that the gravitational field is usually caused by 'big mass', M
- Therefore, the potential energy E is given by:
E = mVg
- Substituting the equation for gravitational potential Vg gives the equation for the gravitational potential energy E between two masses M and m:
- Where:
- G = Universal Gravitational Constant (N m2 kg–2)
- M = mass causing the field (kg)
- m = mass moving within the field of M (kg)
- r = distance between the centre of m and M (m)
Calculating Changes in Gravitational Potential Energy
- When a mass is moved against the force of gravity, work is required
- This is because gravity is attractive, therefore, energy is needed to work against this attractive force
- The work done (or energy transferred) ∆W to move a mass m between two different points in a gravitational potential ∆V is given by:
∆W = m ∆V
- Where:
- ∆W = work done or energy transferred (J)
- m = mass (kg)
- ∆V = change in gravitational potential (J kg-1)
- This work done, or energy transferred, is the change in gravitational potential energy (G.P.E) of the mass
- When ∆V = 0, then the the change in G.P.E = 0
- The change in G.P.E, or work done, for an object of mass m at a distance r1 from the centre of a larger mass M, to a distance of r2 further away can be written as:
- Where:
- M = mass that is producing the gravitational field (e.g., a planet) (kg)
- m = mass that is moving in the gravitational field (e.g., a satellite) (kg)
- r1 = initial distance of m from the centre of M (m)
- r2 = final distance of m from the centre of M (m)
- Work is done when an object in a planet's gravitational field moves against the gravitational field lines, i.e., away from the planet
- This is, again, because gravity is attractive
- Therefore, energy is required to move against gravitational field lines
Gravitational potential energy increases as a satellite leaves the surface of the Moon
Worked example
A spacecraft of mass 300 kg leaves the surface of Mars to an altitude of 700 km.Calculate the work done by the spacecraft.Radius of Mars = 3400 km
Mass of Mars = 6.40 × 1023 kg
Step 1: Write down the work done (or change in G.P.E) equation
Step 2: Determine values for r1 and r2
r1 is the radius of Mars = 3400 km = 3400 × 103 m
r2 is the radius + altitude = 3400 + 700 = 4100 km = 4100 × 103 m
Step 3: Substitute in values
ΔG.P.E = 643.076 × 106 = 640 MJ (2 s.f.)
Examiner Tip
Make sure to not confuse the ΔG.P.E equation with
ΔG.P.E = mgΔh
The above equation is only relevant for an object lifted in a uniform gravitational field (close to the Earth’s surface). The new equation for G.P.E will not include g, because this varies for different planets and is no longer a constant (decreases by 1/r2) outside the surface of a planet.