Effects of gravitational fields

Use this page to revise the following concepts within effects of gravitational fields:

Gravitational field s are a region of space where mass es will experience a force . It is always an attractive force. All masses create gravitational fields. The more massive an object, the larger its gravitational field.
When a large central body, such as a star or planet, produces a gravitational force on a smaller body, such as satellite , the gravitational force is found by.

\[ F = \frac{GMm}{r^2} \]

and hence acceleration is found by:

\[ a = g = \frac{GM}{r^2} \]

Where:

  • \(F\) = gravitational force \((\mathrm{~N})\)
  • \(G\) = gravitational constant \((6.67 \times 10^{-11} \mathrm{~Nm}^2\mathrm{~kg}^{-2})\)
  • \(M\) = central mass \((\mathrm{kg})\)
  • \(m\) = orbiting mass \((\mathrm{kg})\)
  • \(a\) = acceleration \((\mathrm{ms}^{-2})\)
  • \(r\) = distance between the centres of the two masses \((\mathrm{m})\)

This equation can be applied to the forces between any objects but is most often used for forces between an object and its satellite. If we assume a satellite is travelling in a circle around an object then its net force must also be:

\[ F = \frac{mv^2}{r} \]

Since \(v = \frac{2\pi r}{T}\) for circular motion, we get:

\[ F = \frac{m\left(\frac{2\pi r}{T}\right)^2}{r} \]

Equating the expressions for the gravitational and centripetal forces:

\[ \begin{aligned} \frac{GMm}{r^2} &= \frac{m\left(\frac{2\pi r}{T}\right)^2}{r} \\ \frac{GMm}{r^2} &= \frac{4\pi^2mr}{T^2} \\ \frac{r^3}{T^2} &= \frac{GM}{4\pi^2} \end{aligned} \]

Note that for any object with a mass \(M\) its satellites’ motion abides by this equation.

The right-hand side is constant and \(r^3 \propto T^2\).

Importantly, this is independent of satellite mass!

GPE in a gravitational field

The GPE of a mass in a gravitational field varies, according to the inverse square law , depending on the distance of the mass from the object creating the field.

A force-distance graph or a field-distance graph can be used to show the gravitational field strength or force on a body at any distance.

The area under the graph between two points can be used to calculate the change in GPE of an object.

A graph has Gravitational field strength g(N*kg^-1) on the y-axis and radial displacement (r) on the x-axis.

When finding this area, key aspects must be noted:

  • Axis units (is the distance metres or kilometres etc)
  • Vertical axis unit. \(g\) or \(F\)? If it is g then it must be multiplied by the mass of the object.
  • Truncated graph. The graph may be truncated and the full area under the graph must be taken into account, even that which is not visible.

Worked Example

The gravitational force on an object at a distance \(r\) from Earth's centre is graphed below. Determine the change in gravitational potential energy as the object moves from \(20000 \mathrm{~km}\) to \(12000 \mathrm{~km}\) from Earth's centre.

A graph has Force (N) on the y-axis and Distance from centre of Earth in 10^6m on the x-axis. A vertical line at approxing 6.37 on the x-axis is labelled ‘Surface of Earth.’ A curve plots the gravitational force, being near 100 N at the surface of the Earth, and then decreasing sharply in an inverse-square relationship. It crosses the points (10,40) and (20,10).

The radius changes from \(20 \times 10^6 \mathrm{~m}\) to \(12 \times 10^6 \mathrm{~m}\).

Counting the number of squares under the curve between these two values gives around \(6.5\) squares.

Each square has a value of \((2 \times 10^6 \mathrm{~m})(10 \mathrm{~N}) = 2 \times 10^7 \mathrm{~J}\). So the magnitude of the change in GPE is:

\[ |\Delta E_g| = 6.5 \times 2 \times 10^7 = 1.3 \times 10^8 \mathrm{~J} \]

Since the object moves closer to Earth, its gravitational potential energy decreases. Therefore:

\[ \Delta E_g = -1.3 \times 10^8 \mathrm{~J} \]

Worked Example

This change in GPE can be used to determine changes in velocity of an object.
Consider the graph below:

A graph has Gravitational field strength (N*kg^-1) on the y-axis and Distance from Surface of Ceres in 10^5m on the x-axis. An exponential decay line is drawn from the surface of Ceres at 0.28N*kg^-1 to 0N*kg^-1 at 22*10^5m from surface of Ceres.

Let a 10 kg object be dropped from a distance of \(2 \times 10^5 \mathrm{~m}\) above the surface of Ceres. Determine the impact velocity.

The change in GPE is given by the area under the curve. Using rectangles, the area under the curve is about 10 rectangles.

Each rectangle has the area:

\[ \left(2 \times 10^4 \mathrm{~m}\right)\left(0.20 \mathrm{~Nkg}^{-1}\right)=4.0 \times 10^3 \mathrm{~Jkg}^{-1} \]

So the total area is approximately:

\[ 10 \times 4.0 \times 10^3=4.0 \times 10^4 \mathrm{~Jkg}^{-1} \]

For a mass of 10 kg:

\[ \Delta GPE=10 \times 4.0 \times 10^4=4.0 \times 10^5 \mathrm{~J} \]

Once it reaches the surface, this GPE will be entirely converted into KE:

\[ \begin{aligned} \frac{1}{2}mv^2&=4.0 \times 10^5 \\ \frac{1}{2}(10)v^2&=4.0 \times 10^5 \\ v&=\sqrt{8.0 \times 10^4} \approx 283 \mathrm{~ms}^{-1} \end{aligned} \]

Hence, the impact velocity will be approximately \(283 \mathrm{~ms}^{-1}\).