# 5.9.4 Gravitational Potential Energy

## 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:

EmVg

• 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 kg2)
• 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)
• Vchange 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)
• r1initial distance of m from the centre of M (m)
• r2 = final distance of m from the centre of (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 × 10= 640 MJ (2 s.f.)

#### Exam 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. ### Get unlimited access

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