Gravitation – Force and Field Strength

Concept of a Gravitational Field and Force

A gravitational field is a region in which a mass experiences a force.

A mass, \begin{aligned} m \end{aligned} that is present inside a gravitational field experiences a gravitational force. This gravitational field is produced by another mass \begin{aligned} M \end{aligned} .

The amount of force experienced by the mass \begin{aligned} m \end{aligned} is directly proportional to the product of the two masses and indirectly proportional the the square of the separation.

\begin{aligned} F_G=\frac{GM m}{r^2} \end{aligned}

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The separation \begin{aligned} r \end{aligned} is the distance between the centre of mass of the two masses. We often assume that the two masses are point masses if the separation is large relative to the radius of the masses. If the separation is not large, then it is important to use the distance between the centre of mass of the two masses. We should not use the separation between the two surfaces of the masses. Furthermore, one may safely assume that for a uniform sphere, the centre of mass is the centre of the sphere.

There are two regions about the field produced by a mass: the region outside the mass and the region inside. The region outside the mass follows the inverse exponential relationship of 1/r^2.

Inside the mass, the relation is direct proportional to the distance from the centre of the mass. This is because as you proceed nearer to the centre of the mass, there is less mass “below” you. The part of the mass “above” you pulling you “up” is offset by the mass “below” you pulling you down.

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Gravitational Field Strength

Gravitational field strength is often misunderstood. Its definition is

Gravitational field strength at a point is the gravitational force acting on a unit mass at that point.

From the definition,

\begin{aligned} \text{force} &= m \times g\\mg &= \frac{GMm}{r^2}\\g &= \frac{GM}{r^2}\end{aligned}

We can observe that gravitational field strength only depends on the gravity-producing mass and the distance from it. It is not dependent on the test mass.

There is only one value of gravitational field strength at any particular point since we are always comparing the gravitational force on one unit mass. If there are multiple masses creating gravitational fields, the gravitational field strength at any particular point would be the vector sum of all the field strengths due to the the different masses.

Field Strength on the surface of the Earth

When we calculate weight of an object, we always use the formula \begin{aligned} w = mg \end{aligned}

\begin{aligned} g \end{aligned} is referred to as the gravitational field strength (although it is commonly stated as the gravitational acceleration). Near the Earth’s surface, the field strength of the Earth is

\begin{aligned} g &=G\frac{M}{r_\text{Earth radius}^2}\\&=G\frac{M}{6400000^2}\end{aligned}

The field strength at 10 km above the surface would be \begin{aligned} G\frac{M}{6410000^2} \end{aligned}.

The difference among them is negligible. Hence we assume that the gravitational field strength on Earth’s surface is constant.

Activity 1

Access the online PHET simulation.

Objective:
What is the relationship between the force acting on m2 by m1 and the force of m1 on m2?

Task:

  1. Using the default values, observe the force on m2 by m1 and the force on m1 by m2.
    What do you observed?
  2. Change the mass of m1 to another value.
    What do you observed?
  3. Explain.

Activity 2

Access the same simulation as activity 1. Also open this activity sheet to download a copy of the activity on iCloud.

Objective:
What is the relationship between the field strength with distance?

Task:

  1. Change m2 to 1 kg, while m1 remains as 50 kg.
  2. Drag m2 to the 10 m mark, and m1 to the 0 m mark. You may not be able to do this exactly, but an approximate position would be fine.
  3. Record the force on m2 by m1 in the table.
  4. Repeat this for distances 9, 8, 7, 6, 5, 4, 3, 2 and 1 m.
  5. Observe the shape of the graph.

Question:

  1. Will the gravitational field strength ever reaches zero?
  2. At which point(from 0 m to infinity) is the field strength the strongest?

Summary

  1. understand the concept of a gravitational field as an example of a field of force and define gravitational field strength as force per unit mass.
  2. understand that, for a point outside a uniform sphere, the mass of the sphere may be considered to be a point mass at its centre.
  3. recall and use Newton’s law of gravitation in the form \begin{aligned} F=G\frac{Mm}{r^2} \end{aligned}

Review

Question 1

Geostationary satellites are satellites that orbit around the Earth with a period of 24 hours. The satellite would appear as a stationary point relative to an observer on Earth. Calculate the distance above the Earth’s surface of a geostationary satellite. Properties of Earth

Solution

Since the centripetal force of the satellite is due to gravity,

\begin{aligned} \frac{GMm}{r^2} &= mr\omega ^2\\\frac{GM}{r^2} &= r \omega ^2 \\r &= \sqrt[3]{\frac{GM}{w}} \end{aligned}

\omega is known, since the geostationary satellite must make one orbit in one day,

\begin{aligned} \omega &= \frac{2 \pi}{24 \times 3600}\\&= 7.3\times 10^{-5} \text{ rad s}^{-1}\\r&=42.1\times 10^6\text{ m}\end{aligned}

Hence, the distance above Earth’s surface is

\begin{aligned} d &= (42.1 - 6.38) \times 10^6 \text{ m}\\ &= 35.7 \times 10^3 \text{ km} \end{aligned}

It is interesting to note that the distance of Moon from Earth is about 370 \times 10^3 \text{ km}. Hence, a geostationary satellite is about 10% of the distance from Earth to the Moon.