Physics (9702) | Gravitational field of a point mass

A-Level Physics (9702) Study Notes: Gravitational Field of a Point Mass

Hello future physicist! This chapter dives deep into the forces that shape the cosmos—gravity! We are moving beyond the simple \(F=mg\) you learned previously and exploring gravity on a much grander scale, where masses are far apart, like planets and stars.

Don't worry if the formulas look complicated; we will break down each concept using simple steps and analogies. By the end of this chapter, you'll understand why gravitational fields are defined the way they are, and why certain things (like potential) have a weird negative sign!

---

1. The Gravitational Force: A Cosmic Hug

Before defining the field, we must recall the force that creates it: the Gravitational Force, described by Newton's Law of Gravitation.

1.1 Newton's Law of Gravitation (Review)

This law describes the attractive force (\(F\)) between any two point masses, \(m_1\) and \(m_2\), separated by a distance \(r\):

\(F = \frac{G m_1 m_2}{r^2}\)

• \(G\) is the Gravitational Constant (\(6.67 \times 10^{-11} \, \text{N m}^2 \text{ kg}^{-2}\)). It's just a proportionality constant that makes the units work out.
• \(r\) is the distance between the centres of the masses.
• Point Mass Assumption: When dealing with large, uniform spherical objects (like Earth or the Sun), we can simplify our calculations by treating the entire mass of the sphere as if it were concentrated at its centre. This is incredibly useful!

Key Takeaway: The force of gravity is always attractive and decreases rapidly as the distance increases (it follows an Inverse Square Law, proportional to \(1/r^2\)).

---

2. Gravitational Field Strength (\(g\)) for a Point Mass

A field is simply a region of space where a mass experiences a force. The field strength tells us how strong that force is at any specific point.

2.1 Definition of Gravitational Field Strength

The Gravitational Field Strength (\(g\)) at a point is defined as the force per unit mass exerted on a small test mass placed at that point.

\(g = \frac{F}{m}\)

The unit for \(g\) is Newtons per kilogram (\(\text{N kg}^{-1}\)). Wait, isn't that the same as acceleration? Yes! For small changes in height near the Earth, \(g\) is also known as the acceleration of free fall (\(\text{m s}^{-2}\)).

2.2 Deriving and Using \(g = GM/r^2\)

We can combine the definition of \(g\) with Newton's Law of Gravitation to find the formula for the field strength generated by a single mass \(M\).

Step-by-Step Derivation:

1. Start with the definition: \(g = \frac{F}{m}\). (Where \(m\) is the test mass).
2. Substitute Newton's Law for \(F\), where \(m_1=M\) (the central mass) and \(m_2=m\) (the test mass):
   \(g = \frac{(G M m / r^2)}{m}\)
3. The test mass \(m\) cancels out!
   \(g = \frac{G M}{r^2}\)

The formula for the gravitational field strength due to a point mass \(M\) is:

\(g = \frac{G M}{r^2}\)

This equation is vital. It shows that the field strength only depends on:

• \(G\) (constant)
• \(M\) (the mass creating the field, e.g., the Earth)
• \(r\) (the distance from the centre of \(M\))

Important Note: Gravitational Field Strength (\(g\)) is a Vector Quantity. It always points inwards, towards the centre of the mass \(M\). We represent gravitational fields using Field Lines, which show the direction of the force on a mass. For a point mass, these lines are radial (like spokes on a wheel), pointing inwards.

2.3 Why is \(g\) Constant Near Earth?

Near the surface of the Earth, we often assume \(g \approx 9.81 \, \text{N kg}^{-1}\) is constant. Why is this justified?

The Earth's radius (\(R_E\)) is about \(6400 \, \text{km}\). If you climb a mountain 1 km high, your distance \(r\) from the centre changes from \(R_E\) to \(R_E + 1 \, \text{km}\).

The change in \(r\) is very small compared to the huge value of \(R_E\). Since \(g\) depends on \(1/r^2\), the resulting change in \(g\) is negligible. Therefore, for small changes in height (like laboratory experiments or walking up a hill), \(g\) is treated as approximately constant.

---

3. Gravitational Potential (\(\phi\)): The Energy Map

In addition to force (a vector), we also need to understand the energy associated with the field. This brings us to Gravitational Potential, a scalar quantity.

3.1 Defining Gravitational Potential

The Gravitational Potential (\(\phi\)) at a point is defined as the work done per unit mass in bringing a small test mass from infinity to that point.

Wait, infinity? Yes! In physics, "infinity" means a point so far away that the gravitational force from the mass \(M\) is effectively zero. We define the gravitational potential at infinity (\(r = \infty\)) to be zero (\(\phi_{\infty} = 0\)).

The Analogy: The Energy Cost Map
Imagine gravity is a deep hole.

• When you are infinitely far away, you are on the flat ground (Potential = 0).
• As you move closer to the mass \(M\) (the hole), you are moving downhill. The field does the work for you.
• Since you move towards the mass without needing external energy input, the work done (and thus the potential) must be negative.

Therefore, gravitational potential \(\phi\) is always negative in a gravitational field, because gravity is attractive.

3.2 The Formula for Gravitational Potential

The gravitational potential (\(\phi\)) at a distance \(r\) from a point mass \(M\) is:

\(\phi = - \frac{G M}{r}\)

The unit for gravitational potential is Joules per kilogram (\(\text{J kg}^{-1}\)).

Did you know? Gravitational potential depends only on \(1/r\), not \(1/r^2\). This is a general feature of potentials derived from inverse square law forces. Since potential is simpler mathematically (it’s a scalar), it is often easier to use potential rather than field strength for complex calculations!

3.3 Gravitational Potential Difference

In many scenarios, we are more interested in the Gravitational Potential Difference (\(\Delta \phi\)) between two points, A and B. This is the work done per unit mass moving a mass from point A to point B.

\(\Delta \phi = \phi_B - \phi_A\)

If \(\Delta \phi\) is positive, external work was done on the mass (you lifted it out of the hole). If \(\Delta \phi\) is negative, the field did work on the mass (it fell deeper into the hole).

---

4. Gravitational Potential Energy (\(E_p\))

Potential energy is simply the energy stored by a mass due to its position in the field. This concept connects potential (\(\phi\)) back to energy (\(E_p\)).

4.1 Defining Gravitational Potential Energy

The Gravitational Potential Energy (\(E_p\)) of a mass \(m\) at a specific point in a gravitational field is the work done (by an external force) to bring that mass \(m\) from infinity to that point.

Since \(\phi\) is the work done per unit mass, the total potential energy is just the potential multiplied by the mass \(m\):

\(E_p = m \phi\)

Substituting the formula for \(\phi\):

\(E_p = - \frac{G M m}{r}\)

The unit for \(E_p\) is Joules (\text{J}).

4.2 Understanding the Negative Energy

Like potential, \(E_p\) is also always negative. This tells us that the two masses \(M\) and \(m\) are bound together by gravity.

• To separate them completely (move \(m\) to infinity), you would need to supply a positive amount of energy equal to \(\frac{G M m}{r}\).
• The closer the objects are (smaller \(r\)), the more negative \(E_p\) is, meaning the more tightly bound they are.

Think of it like money: If you have \(-100 \text{ J}\) of potential energy, you are in debt to the field. You need to pay \(+100 \text{ J}\) to reach zero energy (infinity).

You've tackled the most abstract part of gravitational fields! Remember, the core difference lies in the definition: force deals with vectors (\(1/r^2\)) and potential deals with energy (scalars, \(1/r\), and negative signs!). Keep practicing the application of these formulas, and you will master this topic!