# Gravitational Potential Energy & Work Done in Gravitational Field

This topic is part of the HSC Physics syllabus under the section Motion in Gravitational Fields.

### HSC Physics Syllabus

• Derive quantitatively and apply the concepts of gravitational force and gravitational potential energy in radial gravitational fields to a variety of situations, including but not limited to:

– total potential energy of a planet or satellite in its orbit U=-(GMm)/r
– total energy of a planet or satellite in its orbit U+K=-(GMm)/(2r)
– energy changes that occur when satellites move between orbits

### What is Gravitational Potential Energy?

All masses experience gravitational force when they are within a gravitational field

All masses possess gravitational potential energy due to its position in a gravitational field. It is also defined as the potential amount of work done by the force of gravity over a certain displacement.

U=mgh

However, as the gravitational force varies with distance from the centre of the mass that produces the gravitational field, a new and more accurate formula is:

U=-(GMm)/r

where:
• G is the universal gravitational constant = 6.67 xx 10^{-11}
• M is the mass that produces the gravitational field in kg
• m is the mass for which gravitational potential energy is calculated in kg
• r is the distance between the centres of the two masses in metres (m)

 U = mgh U = -\frac{GMm}{r} Definition Energy required to lift an object of mass m by height h Energy required to move a mass from a point of infinity to a position within the gravitational field that is distance r from the source of gravity Gravitational field g Assumed to be constant Accounts for the variability of field strength When to use When displacement is small e.g. near surface of Earth When displacement is large e.g. in space or between orbits

### What is Gravitational Potential Energy Negative?

The negative sign in the gravitational potential energy equation is crucial. It suggests that the potential energy is zero when the distance r is infinitely large, meaning the object is free from the influence of the gravitational field. As the object moves closer to the massive body (decreasing r), the potential energy becomes more negative, indicating an increase in the strength of the gravitational field.

• Gravitational potential energy is inversely proportional to orbital radius.

### Work Done in a Gravitational Field

The work done in a gravitational field equals to the change in an object's gravitational potential energy:

W = DeltaU = U_f - U_i

W = -(GMm)/(r_f)+(GMm)/(r_i)

#### Work Done by Gravity

When an object is in a gravitational field, force of gravity does work on the object, bringing it closer to the mass responsible for the field. As r decreases, u decreases and becomes more negative.

#### Work Done Against Gravity

To move an object in a gravitational field, work must be done against the gravitational force. The work done to move an object from one point to another in a gravitational field changes its gravitational potential energy. If an object is moved farther from a massive body, work is done to overcome the gravitational field, thereby increasing the potential energy (making it less negative).

Work done to move an object against gravity equals to change in potential energy of the object during the movement.

### Calculation Example

A satellite with a mass of 200 kg maintains its orbit at an altitude of 300 km above the Earth’s surface. The Earth has a radius of 6.371 x 106 m and a mass of 6.0 x 1024 kg.

(a) Determine the gravitational potential energy of the satellite at this altitude.

(b) Calculate the work done to move this satellite to an altitude of 4000 km above Earth’s surface.

Solution to part (a):

$$U = -\frac{GMm}{r}$$

Note that r = Earth's radius + altitude:

$$U = -\frac{(6.67 \times 10^{-11})(6.0 \times 10^{24})(200)}{6.371 \times 10^6 + 300 \times 10^3}$$

$$U = - 1.2 \times 10^{10} \text{ J}$$

Solution to part (b):

﻿﻿In this instance, work is done against gravity which will increase the satellite's gravitational potential energy:

W = -(GMm)/(r_f)+(GMm)/(r_i)

$$W = -\frac{(6.67 \times 10^{-11})(6.0 \times 10^{24})(200)}{6.371 \times 10^6 + 4000 \times 10^3} + \frac{(6.67 \times 10^{-11})(6.0 \times 10^{24})(200)}{6.371 \times 10^6 + 300 \times 10^3}$$

$$W = 4.3 \times 10^9 \text{ J}$$

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