Physics 9702/43 — May/June 2013

Background Concept

Newton’s law of gravitation describes the gravitational force between two point masses (or spherically symmetric masses treated as point masses at their centres).

The magnitude is

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

where:

• $G$ is the gravitational constant,
• $m_1$ and $m_2$ are the masses,
• $r$ is the separation between their centres.

The force is always attractive and acts along the line joining the centres.

Understanding the Question

The question asks you to state Newton’s law of gravitation (3 marks). To gain full credit, you usually need:

• the correct inverse-square formula,
• the idea that it acts along the line joining the masses,
• and that it is attractive.

Approach

Quote the law in words and/or with the formula, making sure to include the inverse-square dependence and direction.

Step-by-Step Reasoning

1. Write the correct expression with $G$ and $r^2$ in the denominator.
2. Add the direction: along the line joining the centres.
3. State “attractive” to distinguish from other inverse-square forces that can repel.

Key Takeaways

• Gravitational force follows an inverse-square relationship.
• It acts along the line joining centres and is always attractive.

Common Mistakes

• Writing $F = Gm_1m_2/r$ (missing the square).
• Forgetting to mention direction/attractive nature.

Things to Be Careful About

• $r$ is centre-to-centre separation.
• Use correct symbols and avoid mixing with electric force constants.

Background Concept

Gravitational field strength is defined by

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

For a mass $m$ in the gravitational field of a (spherically symmetric) mass $M$ a distance $r$ away,

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

Divide by $m$ to get the field strength due to $M$:

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

So $g$ follows an inverse-square relationship with distance from the source mass.

Understanding the Question

You are given orbital radii of Earth and Jupiter about the Sun. You are asked for the ratio

$$\frac{g_{\text{Earth orbit}}}{g_{\text{Jupiter orbit}}}$$

Both field strengths are due to the same Sun, so $G$ and $M$ should cancel in the ratio.

Approach

1. Use Newton’s law to write $g = GM/r^2$.
2. Form the ratio $g_E/g_J$.
3. Substitute the two orbital radii and square the distance ratio.

Step-by-Step Reasoning

1. Start with Newton’s law:

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

1. Convert force into field strength:

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

1. Write the ratio:

$$\frac{g_E}{g_J} = \frac{GM/R_E^2}{GM/R_J^2} = \frac{R_J^2}{R_E^2} = \left(\frac{R_J}{R_E}\right)^2$$

1. Substitute values (units cancel because it is a ratio):

$$\left(\frac{7.78 \times 10^8}{1.50 \times 10^8}\right)^2 = (5.19)^2 = 26.9$$

So the ratio is about $27$.

Interpretation: the Sun’s gravitational field at Earth’s orbit is about 27 times stronger than at Jupiter’s orbit because Earth is much closer.

Key Takeaways

• For the same central mass, $g \propto 1/r^2$.
• Ratios are powerful because constants cancel.

Common Mistakes

• Inverting the ratio and giving $\approx 1/27$.
• Forgetting to square the distance ratio.
• Mixing up the radii (using $R_E/R_J$ instead of $R_J/R_E$ when forming $g_E/g_J$).

Things to Be Careful About

• The ratio is dimensionless; units can be left as km because they cancel.
• Keep at least 3 significant figures in intermediate steps before rounding.

Background Concept

For uniform circular motion of radius $R$ and speed $v$, the required centripetal acceleration is

$$a = \frac{v^2}{R}$$

The required centripetal force on an orbiting mass $m$ is therefore

$$F_c = ma = \frac{mv^2}{R}$$

For a circular orbit around a much more massive body (the Sun), gravity supplies this centripetal force.

Newton’s law of gravitation gives the gravitational force between the Sun (mass $M$) and the Earth (mass $m$):

$$F_g = \frac{GMm}{R^2}$$

Also, the orbital speed is related to period $T$ by distance traveled in one orbit:

$$v = \frac{2\pi R}{T}$$

Understanding the Question

You are told the Earth’s orbital period is $T$ and the orbital radius is $R$. You must show that the Sun’s mass is

$$M = \frac{4\pi^2 R^3}{GT^2}$$

This is essentially deriving the standard circular-orbit result by setting “gravity = centripetal”.

Approach

1. Write the gravitational force on the Earth due to the Sun.
2. Write the centripetal force needed for circular motion.
3. Equate them (gravity provides centripetal force).
4. Substitute $v = 2\pi R/T$ and rearrange to make $M$ the subject.

Step-by-Step Reasoning

1. Gravitational force magnitude:

$$F_g = \frac{GMm}{R^2}$$

1. Required centripetal force for circular orbit:

$$F_c = \frac{mv^2}{R}$$

1. For a stable circular orbit, the net inward force must equal the required centripetal force, so

$$\frac{GMm}{R^2} = \frac{mv^2}{R}$$

1. Cancel $m$ (important: the orbit does not depend on the mass of the orbiting object):

$$\frac{GM}{R^2} = \frac{v^2}{R}$$

1. Substitute the speed-period relation:

$$v = \frac{2\pi R}{T} \Rightarrow v^2 = \left(\frac{2\pi R}{T}\right)^2 = \frac{4\pi^2 R^2}{T^2}$$

1. Put into the equation:

$$\frac{GM}{R^2} = \frac{1}{R}\cdot \frac{4\pi^2 R^2}{T^2} = \frac{4\pi^2 R}{T^2}$$

1. Rearrange for $M$:

$$M = \frac{4\pi^2 R^3}{GT^2}$$

Key Takeaways

• Circular orbit condition: gravitational force provides centripetal force.
• Use $v = 2\pi R/T$ to connect circular motion to period.
• Mass of the orbiting object cancels.

Common Mistakes

• Using $F = mv/R$ instead of $F = mv^2/R$.
• Forgetting to substitute $v = 2\pi R/T$ (or using $v = 2\pi/T$ without $R$).
• Algebra slip leading to $R^2$ instead of $R^3$ in the final expression.

Things to Be Careful About

• Ensure $R$ is the separation between centres (Sun to Earth).
• Keep $T$ in seconds if you later use the formula numerically.
• This derivation assumes a circular orbit (or a good circular approximation).

Background Concept

For an object in circular orbit about a central mass $M$,

$$M = \frac{4\pi^2 R^3}{GT^2}$$

This comes from combining:

• centripetal motion ($v = 2\pi R/T$ and $F_c = mv^2/R$),
• Newtonian gravitation ($F_g = GMm/R^2$).

To use it numerically, all quantities must be in SI units: $R$ in m, $T$ in s, $G$ in SI.

Understanding the Question

You are given:

• $T = 3.16 \times 10^7\ \text{s}$,
• $R = 1.50 \times 10^8\ \text{km}$ (from the table),

and asked to calculate the Sun’s mass in kg.

Approach

1. Convert the orbit radius from km to m.
2. Substitute $R$, $T$, and $G = 6.67 \times 10^{-11}\ \text{N m}^2\ \text{kg}^{-2}$ into the formula.
3. Calculate carefully in standard form and round appropriately.

Step-by-Step Reasoning

1. Unit conversion:

$$1\ \text{km} = 10^3\ \text{m}$$

So

$$R = 1.50 \times 10^8\ \text{km} = 1.50 \times 10^8 \times 10^3\ \text{m} = 1.50 \times 10^{11}\ \text{m}$$

1. Substitute into

$$M = \frac{4\pi^2 R^3}{GT^2}$$

Compute the powers first:

$$R^3 = (1.50 \times 10^{11})^3 = 3.375 \times 10^{33}$$

and

$$T^2 = (3.16 \times 10^7)^2 = 9.99 \times 10^{14}$$

Now evaluate:

$$M = \frac{4\pi^2 (3.375 \times 10^{33})}{(6.67 \times 10^{-11})(9.99 \times 10^{14})}$$

Since $4\pi^2 \approx 39.48$,

$$\text{numerator} \approx 39.48 \times 3.375 \times 10^{33} \approx 1.33 \times 10^{35}$$

and

$$\text{denominator} \approx 6.67 \times 9.99 \times 10^{(-11+14)} \approx 6.66 \times 10^4$$

So

$$M \approx \frac{1.33 \times 10^{35}}{6.66 \times 10^4} \approx 2.0 \times 10^{30}\ \text{kg}$$

Key Takeaways

• Always convert to SI before substitution.
• Standard form arithmetic is easiest if you separate powers of ten from decimal factors.
• The result is of order $10^{30}\ \text{kg}$, which is the expected scale for the Sun.

Common Mistakes

• Forgetting to convert km to m (would make $M$ smaller by a factor of $10^9$ because of $R^3$).
• Squaring $T$ incorrectly.
• Rounding too early and losing accuracy.

Things to Be Careful About

• Because $R$ is cubed, a small unit error becomes huge.
• Quote the final mass to 2–3 significant figures, consistent with the given data ($1.50$ and $3.16$ are 3 s.f.).