Physics 9702/41 — May/June 2017

Background Concept

For an object to move in a circle of radius $r$ at constant speed $v$, it must have an inward (centripetal) acceleration

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

This acceleration is caused by a resultant inward force (centripetal force)

$$F = ma = m\frac{v^2}{r}.$$

For a satellite orbiting a planet, the natural inward force available is the gravitational force,

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

where $M$ is the planet’s mass and $m$ is the satellite’s mass.

Understanding the Question

The question asks for a qualitative explanation of how a satellite can stay in a circular orbit. The key idea is that the satellite is not “supported”; instead it is constantly accelerating towards the planet due to gravity while moving sideways fast enough to keep missing the surface.

Approach

State (i) the direction of the satellite’s velocity (tangential to the orbit) and (ii) the direction and role of gravity (towards the centre providing centripetal force). A brief “continuous free fall” description completes the explanation.

Step-by-Step Reasoning

1. In a circular orbit, the satellite’s instantaneous velocity is along the tangent to the circle.
2. Gravity acts along the line joining satellite to the planet’s centre, i.e. radially inwards.
3. This inward gravitational force provides the required centripetal force, so the satellite has inward centripetal acceleration while keeping (approximately) constant speed.
4. Because it is continually accelerating towards the planet but also moving sideways, it continually “falls around” the planet rather than straight into it.

Key Takeaways

• Circular motion needs a resultant force towards the centre.
• In orbit, gravity is that centripetal force.
• The satellite’s velocity is tangential; its acceleration (and force) is radial.

Common Mistakes

• Saying “no forces act” on the satellite (gravity must act).
• Confusing tangential velocity with inward acceleration (they are perpendicular).
• Claiming the satellite is in equilibrium (it is not: resultant force is non-zero).

Things to Be Careful About

• Use the phrase “centripetal force” correctly: it is not an extra force, but the name for the resultant inward force.
• Make clear the direction: gravity acts towards the centre of the planet.

Background Concept

For a body of mass $m$ moving in a circular orbit of radius $r$ with angular speed $\omega$, the required centripetal acceleration is

$$a = r\omega^2.$$

So the required centripetal force is

$$F_c = ma = mr\omega^2.$$

For an orbit under gravity, the attractive gravitational force between the planet (mass $M$) and the orbiting body (mass $m$) is

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

In a stable circular orbit, gravity provides exactly the centripetal force:

$$F_g = F_c.$$

Also, density relates to mass and volume. For a uniform sphere of radius $R$ and mean density $\rho$,

$$M = \rho \times \frac{4}{3}\pi R^3.$$

Understanding the Question

We are told:

• Earth radius is $R$ and mean density is $\rho$ (unknown).
• Moon orbits in a circle of radius $nR$ from Earth’s centre.
• Period of orbit is $T$.

We must show that $\rho$ can be written purely in terms of $n$, $T$ and constants, ending with

$$\rho = \frac{3\pi n^3}{GT^2}.$$

This indicates that the Moon’s mass $m$ should cancel out, and the Earth’s mass $M$ should be replaced using $\rho$.

Approach

1. Set gravitational force equal to centripetal force for the Moon.
2. Replace angular speed with $\omega = 2\pi/T$.
3. Replace Earth mass with $M = (4/3)\pi R^3\rho$.
4. Use $r = nR$ and rearrange to make $\rho$ the subject.

Step-by-Step Reasoning

1. For the Moon (mass $m$) in circular orbit of radius $r$:

$$\frac{GMm}{r^2} = mr\omega^2.$$

1. Cancel $m$ (important: the orbital period does not depend on the Moon’s mass for this ideal model):

$$\frac{GM}{r^2} = r\omega^2.$$

1. Substitute $\omega = 2\pi/T$:

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

1. Multiply both sides by $r^2$:

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

1. Replace $M$ with the Earth’s mass in terms of density:

$$M = \frac{4}{3}\pi R^3\rho.$$

So

$$G\left(\frac{4}{3}\pi R^3\rho\right) = r^3\frac{4\pi^2}{T^2}.$$

1. Use $r = nR$, so $r^3 = n^3R^3$:

$$G\left(\frac{4}{3}\pi R^3\rho\right) = n^3R^3\frac{4\pi^2}{T^2}.$$

1. Cancel common factors $4\pi R^3$:

$$\frac{G\rho}{3} = \frac{\pi n^3}{T^2}.$$

1. Rearrange:

$$\rho = \frac{3\pi n^3}{GT^2}.$$

Key Takeaways

• Circular orbit condition: gravitational force supplies centripetal force.
• Use $\omega = 2\pi/T$ to connect orbit period and dynamics.
• Replace mass of a uniform sphere with density (\times) volume.

Common Mistakes

• Using $r = nR$ but forgetting to cube it when $r^3$ appears.
• Not cancelling the Moon’s mass $m$.
• Mixing up $v^2/r$ with $r\omega^2$ and making algebra errors.
• Using $M = \frac{4}{3}\pi r^3\rho$ with the wrong radius (it must be Earth’s radius $R$, not orbital radius $r$).

Things to Be Careful About

• $T$ must be in seconds if you later use SI units numerically.
• Keep track of what each radius represents: Earth’s radius $R$ vs orbital radius $nR$.
• The result is independent of the Moon’s mass only because we assumed an isolated two-body system and a circular orbit with $m \ll M$ (standard exam model).

Background Concept

From part (b), the Earth’s mean density is related to the Moon’s orbital period and orbital radius by

$$\rho = \frac{3\pi n^3}{GT^2},$$

where $n$ is the orbital radius in units of Earth radii ($r=nR$), $G$ is the gravitational constant, and $T$ must be in seconds for SI consistency.

Understanding the Question

You are given:

• $R = 6.38\times 10^3\ \text{km}$
• Moon’s orbital radius $r = 3.84\times 10^5\ \text{km}$
• Orbital period $T = 27.3$ days

You must:

1. Find $n = r/R$.
2. Convert $T$ into seconds.
3. Substitute into $\rho = 3\pi n^3 /(GT^2)$ and give $\rho$ in $\text{kg m}^{-3}$.

Approach

• Compute the ratio $n$ using the distances in the same units (km cancels).
• Convert days to seconds using $1\ \text{day} = 86400\ \text{s}$.
• Evaluate the expression carefully, especially $n^3$ and $T^2$.

Step-by-Step Reasoning

1. Compute $n$:

$$n = \frac{r}{R} = \frac{3.84\times 10^5}{6.38\times 10^3} = 60.2.$$

Using km for both is fine because the units cancel in the ratio.

1. Convert the period to seconds:

$$T = 27.3\times 86400 = 2.35872\times 10^6\ \text{s} \approx 2.36\times 10^6\ \text{s}.$$

1. Substitute into the density formula:

$$\rho = \frac{3\pi (60.2)^3}{(6.67\times 10^{-11})(2.36\times 10^6)^2}.$$

1. Handle powers and large numbers systematically:

• $n^3 \approx 2.18\times 10^5$
• $T^2 \approx 5.57\times 10^{12}$
• $G T^2 \approx 3.71\times 10^2$

So

$$\rho \approx \frac{2.05\times 10^6}{3.71\times 10^2} \approx 5.55\times 10^3\ \text{kg m}^{-3}.$$

Key Takeaways

• Ratios like $n = r/R$ are unit-independent as long as both distances use the same unit.
• Always convert periods to seconds before using formulas with $G$.
• Be careful with powers: the formula involves $n^3$ and $T^2$.

Common Mistakes

• Forgetting to convert days to seconds.
• Cubing $n$ incorrectly or using $n^2$ by accident.
• Using $R$ and $r$ in different units when finding $n$.
• Rounding too early and losing accuracy.

Things to Be Careful About

• Use SI for $G$ ($6.67\times 10^{-11}\ \text{N m}^2\ \text{kg}^{-2}$), so time must be in seconds.
• Quote the final answer to a sensible number of significant figures (typically 3 s.f. here, matching the given data).