Physics 9702/41 — May/June 2022

Background Concept

Newton’s law of gravitation describes the gravitational force between two isolated point masses (or spherically symmetric masses treated as if their mass is concentrated at their centres). It states that the magnitude of the force depends on:

• the product of the masses, and
• the inverse square of the distance between their centres.

The constant of proportionality is the gravitational constant $G$.

Understanding the Question

You are asked to state Newton’s law of gravitation. For full credit, you must give both:

• the inverse-square relationship (or the equation), and
• the direction/nature of the force (along the line joining the masses, attractive).

Approach

Write the standard formula for the force between two point masses and add the key words about proportionality and direction.

Step-by-Step Reasoning

1. Two point masses $m_1$ and $m_2$ separated by distance $r$ exert gravitational forces on each other.
2. The force magnitude is:

$$F = \frac{Gm_1m_2}{r^2}$$

1. The forces are equal and opposite (Newton’s third law) and act along the line joining the centres; the interaction is always attractive.

Key Takeaways

• Gravity between point masses follows an inverse-square law.
• The force is along the line joining the centres and is attractive.

Common Mistakes

• Writing $F \propto 1/r$ instead of $1/r^2$.
• Omitting that the force acts along the line joining the masses.
• Forgetting that the force is attractive.

Things to Be Careful About

• Use $r$ as the centre-to-centre separation.
• State (or imply via the formula) that $G$ is a constant of proportionality.

Background Concept

The gravitational field strength $g$ at a point is defined as the force per unit mass experienced by a small test mass placed at that point:

$$g = \frac{F}{m}$$

For a point mass (or spherically symmetric mass) $M$, Newton’s law of gravitation gives the force on a test mass $m$ a distance $r$ away:

$$F = \frac{GMm}{r^2}$$

Understanding the Question

You are asked to use Newton’s law to show the standard expression for field strength around a point mass:

$$g = \frac{GM}{r^2}$$

So you must start with $F = \dfrac{GMm}{r^2}$ and use the definition $g = F/m$.

Approach

• Write Newton’s gravitational force on a test mass $m$ due to $M$.
• Divide by $m$ (because field strength is force per unit mass).
• Simplify to obtain the required expression.

Step-by-Step Reasoning

1. Place a test mass $m$ at distance $r$ from mass $M$. The gravitational attraction is:

$$F = \frac{GMm}{r^2}$$

1. By definition,

$$g = \frac{F}{m}$$

1. Substitute $F$:

$$g = \frac{1}{m}\left(\frac{GMm}{r^2}\right)$$

1. Cancel $m$:

$$g = \frac{GM}{r^2}$$

This matches what you were asked to show.

Key Takeaways

• Field strength $g$ is derived directly from Newton’s force law by dividing by the test mass.
• The $1/r^2$ dependence is inherited from the inverse-square force.

Common Mistakes

• Writing $g = GMm/r^2$ (forgetting to divide by $m$).
• Using Earth’s surface formula $g = GM/R^2$ without recognising $R$ is just a special case of $r$.

Things to Be Careful About

• The test mass must be small enough not to affect the field significantly (so $M$ is treated as the source mass).
• $r$ is measured from the centre of the point mass (or centre of a sphere).

Background Concept

For a spherical body, the gravitational field outside (and at the surface) is the same as if all the mass were concentrated at its centre. The gravitational field strength at distance $r$ from the centre is:

$$g = \frac{GM}{r^2}$$

At the surface, $r$ equals the radius of the body.

Understanding the Question

You are given the Moon’s:

• mass $M = 7.35 \times 10^{22}\ \text{kg}$,
• radius $r = 1.74 \times 10^6\ \text{m}$,
  and you must show that its surface gravitational field strength due to the Moon alone is $1.62\ \text{N kg}^{-1}$.

Approach

Use

$$g = \frac{GM}{r^2}$$

with $r$ taken as the Moon’s radius, substitute the numbers (including $G = 6.67 \times 10^{-11}\ \text{N m}^2\text{kg}^{-2}$), and evaluate.

Step-by-Step Reasoning

1. Start with the field strength formula:

$$g = \frac{GM}{r^2}$$

1. Substitute the Moon’s values:

$$g = \frac{(6.67 \times 10^{-11})(7.35 \times 10^{22})}{(1.74 \times 10^6)^2}$$

1. Handle the powers of ten and the squared radius carefully:

• Numerator: $6.67 \times 7.35 \approx 49.0$ and $10^{-11} \times 10^{22} = 10^{11}$, so numerator $\approx 4.90 \times 10^{12}$.
• Denominator: $(1.74)^2 \approx 3.03$ and $(10^6)^2 = 10^{12}$, so denominator $\approx 3.03 \times 10^{12}$.

1. Divide:

$$g \approx \frac{4.90 \times 10^{12}}{3.03 \times 10^{12}} = 1.62$$

So

$$g = 1.62\ \text{N kg}^{-1}$$

Key Takeaways

• At a planet/moon surface, use $r = R$ in $g = GM/r^2$.
• Squaring the radius is where most numerical mistakes occur.

Common Mistakes

• Forgetting to square $r$.
• Using $r = 1.74 \times 10^6$ but squaring only the $1.74$ and not the $10^6$.
• Omitting the unit $\text{N kg}^{-1}$.

Things to Be Careful About

• Keep enough significant figures in intermediate steps to get $1.62$.
• Remember $\text{N kg}^{-1}$ is equivalent to $\text{m s}^{-2}$, but quote what the question uses.

Background Concept

Gravitational field strength $g$ is a vector: it has magnitude and direction (direction is towards the attracting mass). When more than one mass produces a field, the resultant field is found by vector addition (principle of superposition).

On the straight line joining two masses, the two gravitational field vectors point in opposite directions between the masses, because each points towards its own centre.

Understanding the Question

You must explain (not calculate yet) why there must be a point $X$ on the line between Earth’s and Moon’s centres where the resultant gravitational field strength is zero. That means the field due to Earth at $X$ has the same magnitude as the field due to the Moon at $X$, but opposite direction.

Approach

• Use the idea that fields add vectorially.
• Note that between the bodies the directions oppose.
• Use $g = GM/r^2$ to argue that as you move along the line, one contribution increases while the other decreases, so equality must occur somewhere.

Step-by-Step Reasoning

1. Consider any point on the line between the centres.
2. The Earth’s gravitational field at that point points towards the Earth.
3. The Moon’s gravitational field at that point points towards the Moon.
4. Therefore the two field vectors are opposite in direction.
5. The magnitudes depend on distance via $g = GM/r^2$:
   • closer to the Moon, the Moon’s field is large and the Earth’s is smaller (because you are far from Earth),
   • closer to the Earth, the Earth’s field is large and the Moon’s is very small.
6. Since one decreases while the other increases as you move from Moon to Earth, there must be an intermediate position where they are equal in magnitude; at that point the equal and opposite vectors cancel, giving resultant field strength $0$.

Key Takeaways

• Gravitational fields superpose as vectors.
• Between two masses, the fields oppose in direction.
• Inverse-square dependence guarantees a crossover point where magnitudes match.

Common Mistakes

• Saying the fields “add to zero” without stating they are in opposite directions.
• Thinking the zero-field point must be halfway (it is not, unless masses are equal).
• Confusing field strength with force: the reasoning is for the field at a point, not for motion of the bodies.

Things to Be Careful About

• The point of zero resultant must lie between the masses; outside the interval, both fields point the same way (towards both masses), so they cannot cancel.
• Use the words “equal magnitude” and “opposite directions” for full credit.

Background Concept

At a point on the line between two masses, the gravitational field due to each mass has magnitude

$$g = \frac{GM}{r^2}$$

and direction towards that mass. The resultant field is the vector sum. For the resultant to be zero, the two contributions must have:

• equal magnitudes, and
• opposite directions.

Between Earth and Moon the directions are opposite, so the zero point occurs where the magnitudes are equal.

Understanding the Question

Given:

• Earth mass $M_E = 5.98 \times 10^{24}\ \text{kg}$
• Moon mass $M_M = 7.35 \times 10^{22}\ \text{kg}$
• centre separation $d = 3.84 \times 10^8\ \text{m}$

We must find the distance $x$ from the Moon’s centre to point $X$ where the resultant gravitational field strength is zero.

Approach

1. Let $x$ be the distance from the Moon to $X$; then the distance from the Earth to $X$ is $d-x$.
2. Set the magnitudes of the two fields equal:

$$\frac{GM_M}{x^2} = \frac{GM_E}{(d-x)^2}$$

1. Cancel $G$ and solve for $x$.

Step-by-Step Reasoning

1. Define distances:

   • Moon to $X$: $x$
   • Earth to $X$: $d-x$ where $d = 3.84 \times 10^8\ \text{m}$

2. Write equality of field magnitudes at $X$:

$$\frac{GM_M}{x^2} = \frac{GM_E}{(d-x)^2}$$

1. Cancel $G$ (same constant on both sides):

$$\frac{M_M}{x^2} = \frac{M_E}{(d-x)^2}$$

1. Rearrange to collect $x$ terms. A clean way is to take square roots:

$$\frac{d-x}{x} = \sqrt{\frac{M_E}{M_M}}$$

1. Compute the mass ratio inside the square root:

$$\frac{M_E}{M_M} = \frac{5.98 \times 10^{24}}{7.35 \times 10^{22}} = 81.36$$

So

$$\sqrt{\frac{M_E}{M_M}} = \sqrt{81.36} = 9.02$$

1. Substitute back:

$$\frac{d-x}{x} = 9.02 \Rightarrow d-x = 9.02x$$

1. Solve for $x$:

$$d = 10.02x \Rightarrow x = \frac{d}{10.02} = \frac{3.84 \times 10^8}{10.02}$$ $$x = 3.83 \times 10^7\ \text{m}$$

This is closer to the Moon, as expected because the Moon is much less massive than Earth (so you need to be closer to the Moon for its field to match Earth’s field).

Key Takeaways

• Zero resultant field between two masses occurs where $\dfrac{GM_1}{r_1^2} = \dfrac{GM_2}{r_2^2}$.
• The point lies closer to the smaller mass.
• Using square roots simplifies the algebra for inverse-square equalities.

Common Mistakes

• Using $d+x$ instead of $d-x$ for the Earth-to-$X$ distance.
• Forgetting to cancel $G$ (not wrong, just extra work).
• Assuming $x = d/2$.
• Mixing up which mass goes with which distance (Moon mass must go with distance from Moon).

Things to Be Careful About

• $x$ is measured from the Moon’s centre, not from the Moon’s surface.
• Keep $d$ in metres and use consistent units throughout.
• When taking square roots, use the positive root because distances are positive.