Physics 9702/43 — May/June 2010

Background Concept

For a spherically symmetric body (mass concentrated at its centre), the gravitational potential at distance $r$ from the centre is

$$\phi(r) = -\frac{GM}{r},$$

taking $\phi=0$ at infinity. This is why the graph approaches $0$ as $r$ increases.

Understanding the Question

You are given a graph of gravitational potential $\phi$ against distance $x$ from Earth’s centre, and the Earth’s radius $R = 6.4\times 10^{6}\ \text{m}$.

At the Earth’s surface, the distance from the centre is $x=R$. Reading $\phi$ at $x=R$ lets you solve for Earth’s mass $M$ using $\phi=-GM/R$.

Approach

1. Read the surface potential $\phi(R)$ from the graph.
2. Use $\phi(R) = -GM/R$.
3. Rearrange to get $M = -\phi(R)R/G$.
4. Substitute values carefully (including the negative sign and powers of ten).

Step-by-Step Reasoning

1. From the graph, at $x=R$ the potential is about $-6.3$ on the scale of $10^{7}\ \text{J kg}^{-1}$, so

$$\phi(R) \approx -6.3 \times 10^{7}\ \text{J kg}^{-1}.$$

1. Use the spherical potential formula:

$$\phi(R) = -\frac{GM}{R}.$$

1. Rearrange:

$$M = -\frac{\phi(R)\,R}{G}.$$

1. Substitute $R=6.4\times 10^{6}\ \text{m}$ and $G=6.67\times 10^{-11}\ \text{N m}^2\ \text{kg}^{-2}$:

$$M \approx -\frac{(-6.3\times 10^{7})(6.4\times 10^{6})}{6.67\times 10^{-11}} \approx 6.0\times 10^{24}\ \text{kg}.$$

The sign works out because $\phi$ is negative, but mass must be positive.

Key Takeaways

• For an isolated sphere, $\phi \propto -1/r$.
• Surface potential gives $M$ via $\phi(R)=-GM/R$.
• Care with signs: $\phi$ is negative, $M$ is positive.

Common Mistakes

• Using $r=2R$ or another value instead of $r=R$ for the surface.
• Dropping the negative sign and getting a negative mass.
• Reading the graph value but forgetting the axis scale factor $10^{7}$.

Things to Be Careful About

• $x$ is distance from the centre of Earth.
• Quote $M$ to a sensible number of significant figures consistent with the graph reading (typically 2 s.f.).

Background Concept

If gravitational potential is defined with $\phi(\infty)=0$, then for a mass $m$ at radius $r$ the gravitational potential energy is

$$E_p = m\phi(r).$$

If an object falls from infinity starting from rest, its total mechanical energy per unit mass stays constant:

$$\frac{1}{2}v^2 + \phi = \text{constant}.$$

At infinity, $v=0$ and $\phi=0$, so the constant is $0$:

$$\frac{1}{2}v^2 + \phi = 0 \Rightarrow \frac{1}{2}v^2 = -\phi.$$

This is a very efficient way to find the speed from a known potential.

Understanding the Question

A meteorite starts at rest at infinity and moves towards Earth. You must find its speed when it is a distance $2R$ above Earth’s surface.

“$2R$ above the surface” is an altitude. The gravitational potential graph is plotted against distance from the centre, so you must convert to radius from the centre:

$$r = R + 2R = 3R.$$

Then you need $\phi$ at $r=3R$, and use energy conservation to get $v$.

Approach

1. Convert altitude to distance from Earth’s centre ($r=3R$).
2. Find $\phi(3R)$ from the graph (or from the $1/r$ form).
3. Apply conservation of energy per unit mass: $\tfrac12 v^2 = -\phi(3R)$.
4. Solve for $v$.

Step-by-Step Reasoning

1. Distance from centre:

$$r = 3R.$$

1. From the graph, $\phi(R)\approx -6.3\times 10^{7}\ \text{J kg}^{-1}$. For an isolated sphere $\phi\propto -1/r$, so tripling $r$ makes the potential one third as negative:

$$\phi(3R) = \frac{\phi(R)}{3} \approx -2.1\times 10^{7}\ \text{J kg}^{-1}.$$

1. Use energy per unit mass. At infinity, total specific energy is $0$. So at $3R$:

$$\frac{1}{2}v^2 + \phi(3R)=0.$$

Hence

$$\frac{1}{2}v^2 = -\phi(3R) = 2.1\times 10^{7}.$$

1. Solve for speed:

$$v = \sqrt{2(2.1\times 10^{7})} = \sqrt{4.2\times 10^{7}} \approx 6.5\times 10^{3}\ \text{m s}^{-1}.$$

Key Takeaways

• Convert “above the surface” to radius from the centre before using $\phi(r)$.
• Falling from rest at infinity gives the very useful relation $\tfrac12 v^2 = -\phi$.
• Gravitational potential is negative; the minus sign is essential.

Common Mistakes

• Taking the distance as $2R$ instead of $3R$.
• Using $\Delta \phi$ with the wrong sign (getting negative kinetic energy).
• Treating $\phi$ as if it were gravitational field strength $g$.

Things to Be Careful About

• Use distance from the centre consistently.
• Keep units as $\text{J kg}^{-1}$ for potential; then $v$ comes out in $\text{m s}^{-1}$ because $1\ \text{J kg}^{-1} = 1\ \text{m}^2\ \text{s}^{-2}$.

Background Concept

Gravitational effects from more than one body combine by superposition:

• The net gravitational field at a point is the vector sum of the individual fields.
• The net gravitational potential at a point is the scalar sum of the individual potentials.

So with Earth and Moon present, the meteorite experiences an additional gravitational pull towards the Moon as well as towards Earth.

Understanding the Question

In the earlier parts, Earth was treated as an isolated sphere so the meteorite would accelerate directly towards Earth’s centre (a symmetric situation).

Now the Moon is present. The question asks for two changes to the motion (so typically one about direction/trajectory and one about speed/energy, or two distinct trajectory outcomes).

Approach

Think in terms of:

1. Direction of acceleration: the Moon adds a sideways component of gravitational force, so the meteorite’s velocity direction changes.
2. Speed/energy: the Moon changes the potential energy landscape, so the speed at a given location relative to Earth alone will not be the same.

Step-by-Step Reasoning

• Without the Moon, forces point towards Earth’s centre only, giving a path aimed at Earth.
• With the Moon, there is an additional force towards the Moon. Unless the meteorite is exactly on the Earth–Moon line in a symmetric position, this extra force is not collinear with the Earth’s force.
• Therefore the resultant acceleration is not directly towards Earth’s centre, so the path bends (deflects) towards the Moon.
• Because potential is a scalar that adds, the total gravitational potential at a point is more negative than with Earth alone (especially near the Moon). As the meteorite moves into a region of lower (more negative) potential, it loses more potential energy and gains kinetic energy differently, so its speed profile changes.
• Depending on the geometry, the meteorite could pass closer to the Moon, be pulled into collision with the Moon, or be gravitationally deflected so it misses Earth.

Key Takeaways

• Superposition: fields add as vectors, potentials add as scalars.
• An extra massive body generally causes (i) a change in direction (trajectory) and (ii) a change in speed/energy compared with the isolated-body case.

Common Mistakes

• Saying only “it goes faster” twice (need two distinct changes).
• Claiming the potential graph of Earth alone still applies exactly (it does not when another mass is nearby).
• Ignoring that force is a vector: the key effect is often a sideways deflection.

Things to Be Careful About

• Keep statements qualitative but physically specific (e.g. “deflected towards the Moon” is better than “path changes”).
• If you mention collision/capture, tie it to the Moon’s gravitational attraction (not random chance).