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Gravitational Field

Warning

Most of this is directly copied from Year 4 Physics Chapter 4 notes

Gravitational force is different from tension, friction and contact force (which all require some form of contact). Since gravitational force can be exerted at a distance, we associate it with a "gravitational field". The gravitational field of a mass can be thought of as the region surrounding that mass where another mass will experience a gravitational force.

Representing Gravitational Field

A common way to represent a gravitational field in a region of space is to use field lines. Gravitational field lines are related to the gravitational field according to:

  1. Direction The gravitational field vector \(\vec g\) is tangent to the gravitational field line at each point in space.
  2. Magnitude/Field Strength The magnitude of the gravitational field is proportional to the number of field lines per unit area through a surface perpendicular to the lines. Therefore \(\vec g\) is larger when the lines are close together and smaller when they are further apart.

The gravitational field in the vicinity of a point mass is always directed towards the mass.

The magnitude of the gravitational field is inversely proportional to the square of the distance from the point mass. Therefore, at regions further from the point mass, there are fewer fields lines per unit area.

So how do we assign a number to \(g\)?

Gravitational Field Strength \(\vec g\)

  • The Earth's gravitational field is stronger at points nearer Earth than at points further from Earth
  • The further we are from a amass, the weaker the gravitational force and the weaker the gravitational field
  • The strength of a gravitational field can be different at different points in the field

How do we 'measure' the strength of the gravitational field?

  1. Set up a gravitational field (we need a decently large mass \(M\) to do that)
  2. Place a second mass \(m\) (called the "test mass") in the gravitational field at a distance \(r\) from the center of \(M\). Note this \(m\) must not significantly affect \(M\)'s gravitational field (which means \(m\) must be small)

  3. Determine the force \(F_g\) that \(M\) exerts on \(m\)

    Unfortunately, it wouldn't be enough (from a scientific point of view) to say that \(F_g\) is the strength of the gravitational field of \(M\) at a point of distance \(r\) from the center of \(M\).

    Why?

    Because \(F_g\) would be different if we used a different \(m\) of different mass.

    If Putin from Russia uses a 2.0kg mass and Kim from Korea uses a 4.0kg mass, obviously the force experienced by the 4.0kg mass would be double even if \(M\) and \(r\) were to be kept constant. [sic]

  4. To ensure consistency, we need to have a 'standard second mass \(m\)'. We shall use 1.0kg - a UNIT mass.

Gravitational field strength, \(\vec g\)

The gravitational field strength at a point is defined as the gravitational force per unit mass acting on a small mass placed at the point.

\[g = \frac{F_g}{m}\]

as \(g\) is actually a vector because \(F_g\) is a vector, hence a more accurate formula would be:

\[\vec g = \frac{\vec F_g}{m} \]

\(\vec g\) has the same direction as \(\vec F_g\)

\(\vec g\)'s units is either \(N/kg\) or \(m/s^2\)

Notes for \(g\)

  1. Mass \(m\) is a particle that is placed in the gravitational field. It is not the mass that creates the gravitational field. Hence mass \(m\) is often known as the test mass.
  2. The mass \(M\) that creates the gravitational field is known as the source mass. THe source mass creates the field, NOT the test mass. The test mass is not necessary for the field to be present. Hence gravitational field strength is a property of the point in space
  3. The value \(g\) near the surface of the Earth is usually taken to be \(9.81 N/kg\). This means that gravitational force (weight) of a \(1.0kg\) mass near the Earth's surface is \(9.81N\).

Gravitational field strength = Acceleration due to gravity (free fall)

Why?

  1. \(g\) and acceleration are dimensionally equivalent.

  2. By definition, \(g = \frac{F_g}{m}\). Therefore \(F_g = mg\)

    From Newton's second law we have \(F_{net} = ma\)

    When a body falls freely, the only force acting on it is \(F_g\). Therefore:

    \[\begin{align*}F_g &= F_{net} \\ mg &= ma \\ g &= a\end{align*}\]

Gravitational Field of a Point Mass

If \(m_1 = M\) (the mass of the source mass) and \(m_2 = m\) (the mass of the test mass), then by definition, \(g = \frac{F_g}{m}\) and \(F_g = G\frac{Mm}{r^2}\)

Therefore:

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

Hence the magnitude of the gravitational field strength due to point mass \(M\) varies inversely to the square of the distance, i.e.:

\[g \propto \frac{1}{r^2}\]

g_outside.png g_outside.png

Question

At a point outside the Earth at a distance \(x\) from its centre, the Earth's gravitational field is about \(5.0 N/kg\); at the Earth's surface, the field is about \(10N/kg\). Which one of the following gives the value for the radius of the Earth?

A. \(\frac{x}{5}\), B. \(\frac{x}{2\sqrt{2}}\), C. \(\frac{x}{\sqrt{2}}\), D. \(x\sqrt{2}\)

C (\(\frac{x}{\sqrt{2}}\))

We can rearrange \(g\propto \frac{1}{r^2}\) to \(r \propto \frac{1}{\sqrt{g}}\). Therefore a doubling in \(g\) means the radius is divided by a factor of \(\sqrt{2}\)

Weightlessness and Apparent Weight

True Weightlessness

From the definition of weight as the gravitational force that acts on a body, True weightlessness = NO gravitational force acting on a body

When will a body experience true weightlessness?

  1. Applying \(F_g = \frac{Gm_1m_2}{r^2}\), \(F_g = 0\) when \(r\) is infinitely large.
  2. Since \(F_g\) is actually a vector (\(\vec F_g\)), it can be \(0\) at a point between 2 bodies (\(\vec F_{g1} + \vec F_{g2} = \vec 0\))

Apparent Weightlessness

A body will also 'appear' to have no weight (reading of \(0\) on a weighing scale) when Normal contact force is \(0\)

When will a body experiene apparent weightlessness?

  1. A body that falls freely (no air resistance)
  2. An astronaut orbiting the earth in a space vehicle at constant height