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#### Notes by Category University Engineering

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# Gravitational Fields

A gravitational field is a region where an object experiences a non-contact force because of its and another object’s mass. Gravitational fields are always attractive, and as such the values are always negative. Spherical objects, or things modelled as spheres such as planets and satellites, can be modelled as a point with mass at its centre.

Field Lines map gravitational fields, including their strength. The closer together, the greater the field strength. Gravitational field strength is given by:

g = F/m gravitational field strength = gravitational force / mass

## Newton's Law of Gravitation

Newton’s Law of Gravitation states that the gravitational force experienced by two objects interacting is directly proportional to the product of their masses and inversely proportional to the square of their separation:

F= −GMm/r²

Where M and m are the masses of the two objects, G is the gravitational constant, and r is the separation between the centre of the objects. This means the distance between them, plus their radii.

The gravitational field strength at a certain point from a single point mass is given as the same, divided by one of the masses:

g = −GM/r² Close to the surface of a planet, gravitational field strength can be modelled as a uniform field, and numerically equals the acceleration of free fall (9.81ms-²).

G is the gravitational constant: 6.67 E-11

## Planetary Motion

The motion of planets around their star can be described using Kepler's laws:

1. Each planet moves in an ellipse around the sun, with the sun at one focus.

2. A line joining the sun to a planet will sweep out equal areas in equal times.

3. The square of the period of orbit is directly proportional to the cube of the radius: T² ∝ r³ The time the planet takes to travel between A and B is the same as the time taken between D and C. The shaded areas are the same.

### Kepler's 3rd Law

The circular motion on an orbiting object occurs because there is a constant centripetal force acting on it due to the gravitational force of the focus. Using the equations for circular motion and Newton's law of gravitation we can derive Kepler's third law:

F = mv²/r Centripetal force

F = -GMm/r² Newton's law of gravitation

T = 2πr/v Time period of circular motion

mv²/r = -GMm/r² Equating centripetal force and Newton's law of gravitation

v² = -GM/r m and r cancel

v = √(GM/r) Square-rooting both sides

T = 2πr / √(GM/r) Sub into time period of circular motion

T² = 4π²r² / (GM/r) Rearrange the fractions

T² = 4π²/GM r³ Kepler's 3rd law

Here we can see that the coefficient in Kepler's 3rd Law is 4π²/GM, and that time period is independent of mass of orbital (the M represents the mass of the planet/star being orbited).

### Geostationary Satellites

Geostationary satellites are satellites that orbit above the earth's equator, and have a time period of exactly 24 hours. This means they are always above the same area of land/sea. This makes them incredibly useful for communication, as you do not have to constantly realign transmitters and receivers, both on earth and the satellite.

From Kepler's third law, we can work out the radius of a geostationary satellite:

24 hours = 86,400 seconds

86,400² = 7,464,960,000

7,464,960,000 x GM = 2.99 E24

2.99 E24 / 4π² = 7.57 E22

³√7.57 E22 = 42,000,00 m = 42,000 km

## Gravitational Potential and Energy

Gravitational potential at a point is the work done to move a unit mass from infinity to the point. It is always negative, with it reaching a maximum of 0 at infinity. In a radial field, Gravitational Potential is given as:

V(g) = −GM/r Gravitational potential = - G x mass of object / distance from centre of object

Instead of measuring the potential from infinity, we tend to measure difference in potential as an object gains or loses height, or between two objects. Therefore, you always need to set a zero point, from which to calculate the change in potential.

### Gravitational Potential Energy

When an object is moved, work is done against gravity. The amount of energy required to do this work depends on the mass of the object and the potential difference:

E = m x V(g) Energy = mass x gravitational potential

Combining this with the equation for gravitational potential gives:

E = GMm/r Energy = G x the two masses / the separation Force-distance graphs show how the magnitude of the force sue to the gravitational field changes with distance. Since F = GMm/r², multiplying by r gives the energy equation above. Therefore, the area under the curve is equal to the work done in moving the object.

### Escape Velocity

For something to escape a gravitational field, its kinetic energy must be at least equal and opposite to its gravitational potential energy – this is known as escape velocity:

v = √(2GM/r) escape velocity = Square root of (2 x G x mass of planet / orbital radius)

Note that the mass is of the planet/star causing the gravitational field. This is because the mass of the orbiting object trying to escape cancels:

1/2 mv² + -GMm/r = 0 Total KE + GPE = 0 for escape velocity 1/2 v² + -GM/r = 0 Cancel the m's

1/2 v² = GMm/r Add -GM/r to both sides

v² = 2GM/r Multiply by two

v = √(2GM/r) Square root both sides

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