Physics 9702/43 — October/November 2015

Background Concept

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

• the masses $m_1$ and $m_2$ (bigger masses give a bigger force), and
• the separation $r$ between their centres (greater separation gives a smaller force).

Mathematically,

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

where $G$ is the gravitational constant. The force is always attractive and acts along the line joining the centres of the two masses.

Understanding the Question

You are asked to state the law, so you must give the proportionalities (to $m_1m_2$ and to $1/r^2$), the fact it is attractive, and the direction (along the line joining centres). Writing the formula is usually expected for full credit.

Approach

Provide a clear sentence (or two) describing the relationship and direction, and include the standard equation with correct symbols.

Step-by-Step Reasoning

1. Identify the interaction: gravitational force between two masses.
2. State the dependence on masses: $F \propto m_1m_2$.
3. State the dependence on separation: $F \propto 1/r^2$.
4. Combine these into one expression using the constant of proportionality $G$:

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

1. Add the directional statement: force acts along the line joining the masses and is attractive.

Key Takeaways

• Gravitational force follows an inverse-square law.
• It depends on the product of the two masses.
• It acts along the line joining the centres and is attractive.

Common Mistakes

• Writing $F = Gm/r^2$ (missing one of the masses).
• Forgetting to mention the line of action / that the force is attractive.
• Using $r$ as distance from a surface instead of centre-to-centre separation.

Things to Be Careful About

• Use centre-to-centre separation $r$.
• State both proportionalities and the direction to secure full marks.

Background Concept

A body in uniform circular motion needs a centripetal (centre-seeking) acceleration towards the centre:

$$a = \frac{v^2}{r} = r\omega^2.$$

So the resultant force towards the centre must be

$$F_c = m\frac{v^2}{r} = mr\omega^2.$$

In an orbit, the gravitational attraction between the planet (mass $M$) and the moon (mass $m$) provides this inward force:

$$F_g = \frac{GMm}{r^2}.$$

For a circular orbit of radius $x$, we set $r=x$.

Understanding the Question

You are told moons move in circular orbits and both planet and moon can be treated as point masses. You must show the relationship between orbital radius $x$ and orbital period $T$:

$$GM = \frac{4\pi^2x^3}{T^2}.$$

This is essentially a form of Kepler’s third law for circular orbits around the same central mass $M$.

Approach

1. Write the gravitational force on the moon due to the planet.
2. Write the centripetal force needed for circular motion.
3. Equate them (gravity provides the centripetal force).
4. Replace $v$ using the period: $v = \frac{2\pi x}{T}$.
5. Rearrange to reach the required expression.

Step-by-Step Reasoning

1. Gravitational force magnitude between masses $M$ and $m$ separated by distance $x$:

$$F_g = \frac{GMm}{x^2}.$$

1. Centripetal force needed for a mass $m$ moving at speed $v$ in a circle of radius $x$:

$$F_c = \frac{mv^2}{x}.$$

1. In a stable circular orbit, gravity supplies exactly the centripetal force:

$$\frac{GMm}{x^2} = \frac{mv^2}{x}.$$

1. Express orbital speed using period. In one orbit the moon travels distance $2\pi x$ in time $T$:

$$v = \frac{2\pi x}{T}.$$

1. Substitute for $v$:

$$\frac{GMm}{x^2} = \frac{m}{x}\left(\frac{2\pi x}{T}\right)^2 = \frac{m}{x}\cdot \frac{4\pi^2x^2}{T^2}.$$

1. Cancel $m$ (important: orbital radius/period do not depend on the moon’s mass for a given $M$), then rearrange:

$$GM = \frac{4\pi^2x^3}{T^2}.$$

Key Takeaways

• For circular orbits, set gravitational force equal to centripetal force.
• Use $v = \frac{2\pi r}{T}$ to connect period and radius.
• The orbit relationship depends on the central mass $M$, not the orbiting mass $m$.

Common Mistakes

• Using $F_c = mv^2$ instead of $F_c = mv^2/r$.
• Using diameter instead of radius in $v = 2\pi r/T$.
• Forgetting to cancel $m$ or algebra mistakes when rearranging.
• Mixing $r$ and $x$ inconsistently.

Things to Be Careful About

• $x$ is the centre-to-centre orbital radius.
• The derivation assumes a circular orbit (constant $v$) and that gravity is the only significant force.
• Keep $4\pi^2$ (not $2\pi$) after squaring the speed expression.

Background Concept

From part (b), the orbit relationship for a moon in a circular orbit around a planet of mass $M$ is

$$GM = \frac{4\pi^2x^3}{T^2}.$$

Rearranging into a straight-line form helps connect it to the graph:

$$x^3 = \left(\frac{GM}{4\pi^2}\right) T^2.$$

This matches $y = m x$ (a straight line through the origin) with:

• $y \equiv x^3$
• $x \equiv T^2$
• gradient $m \equiv \frac{GM}{4\pi^2}$ (provided $x^3$ is plotted in SI and $T^2$ in SI).

If the axes are in non-SI units (km, day), you must convert the gradient into $\text{m}^3\ \text{s}^{-2}$ before using it in $GM$.

Understanding the Question

You are given a straight-line graph of $x^3$ (vertical axis) against $T^2$ (horizontal axis) for Neptune’s moons.

• You must read the gradient of the best-fit line from the graph.
• Use the relationship from (b) to find $GM$.
• Then divide by $G$ to obtain $M$, the mass of Neptune.

The graph axes are scaled:

• vertical axis is $x^3 / 10^{14}\ \text{km}^3$
• horizontal axis is $T^2 / \text{day}^2$

So the gradient you read is in $10^{14}\ \text{km}^3\ \text{day}^{-2}$.

Approach

1. Rewrite the equation as $x^3 = kT^2$ so the gradient corresponds to $k$.
2. Find the gradient from two well-separated points on the best-fit line.
3. Convert the gradient into SI units ($\text{m}^3\ \text{s}^{-2}$).
4. Use $k = GM/(4\pi^2)$ (after consistent units) to calculate $GM$.
5. Use $M = (GM)/G$.

Step-by-Step Reasoning

1. Linear form:

$$x^3 = \left(\frac{GM}{4\pi^2}\right)T^2.$$

So the (SI) gradient of a graph of $x^3$ vs $T^2$ would be $GM/(4\pi^2)$.

1. Read two points on the straight line (choose far apart to reduce percentage reading error). From the figure, a suitable estimate is approximately from the origin to about $(0.35, 4.5)$ in the given axis units.

Hence the plotted-number gradient is

$$m_{\text{plot}} \approx \frac{4.5}{0.35} \approx 13.$$

1. Convert the plotted gradient to actual gradient in $\text{m}^3\ \text{s}^{-2}$.

First convert the $x^3$ scale:

$$10^{14}\ \text{km}^3 = 10^{14}\times (10^3\ \text{m})^3 = 10^{23}\ \text{m}^3.$$

So

$$\frac{x^3}{T^2(\text{day}^2)} = 13 \times 10^{23}\ \text{m}^3\ \text{day}^{-2}.$$

Now convert $\text{day}^{-2}$ to $\text{s}^{-2}$.
Because $1\ \text{day} = 86400\ \text{s}$,

$$T^2(\text{day}^2) = \frac{T^2(\text{s}^2)}{86400^2} \quad \Rightarrow \quad \frac{1}{T^2(\text{day}^2)} = \frac{1}{T^2(\text{s}^2)}\,86400^2.$$

So to express the gradient per $\text{s}^2$, divide by $86400^2$:

$$m_{\text{SI}} = \frac{13 \times 10^{23}}{86400^2}\ \text{m}^3\ \text{s}^{-2}.$$

1. Relate gradient to $GM$. Since with SI units

$$m_{\text{SI}} = \frac{GM}{4\pi^2},$$

then

$$GM = 4\pi^2 m_{\text{SI}} = 4\pi^2\left(\frac{13 \times 10^{23}}{86400^2}\right).$$

Numerically, $86400^2 \approx 7.46\times 10^9$, so

$$\frac{13 \times 10^{23}}{86400^2} \approx 1.7 \times 10^{14},$$

and thus

$$GM \approx 39.5\times 1.7\times 10^{14} \approx 6.9 \times 10^{15}\ \text{m}^3\ \text{s}^{-2}.$$

1. Finally,

$$M = \frac{GM}{G} \approx \frac{6.9 \times 10^{15}}{6.67\times 10^{-11}} \approx 1.0 \times 10^{26}\ \text{kg}.$$

Key Takeaways

• A straight-line plot of $x^3$ vs $T^2$ for moons orbiting the same planet has gradient related to $GM$.
• The gradient must be in SI units before using it in gravitational equations.
• Reading far-apart points on the best-fit line improves the accuracy of the gradient.

Common Mistakes

• Using a single data point rather than the best-fit line to find the gradient.
• Forgetting that the y-axis is scaled by $10^{14}\ \text{km}^3$.
• Converting km to m incorrectly for volumes (must cube the conversion factor).
• Treating $T^2$ in day$^2$ as if it were already in s$^2$.
• Using $GM = m$ instead of $GM = 4\pi^2 m$.

Things to Be Careful About

• For volumes: $1\ \text{km}^3 = 10^9\ \text{m}^3$.
• For time-squared: multiply or divide by $86400^2$ consistently (here you divide the per-day$^2$ gradient by $86400^2$ to get per-s$^2$).
• Quote the mass to a sensible number of significant figures consistent with graph-reading uncertainty (typically 2–3 s.f.).