2019 A/L Physics Past Paper · Question 05 (English Medium)

Answer: (2) Graph (2)

What the question asks

We want:

The graph of the potential $V(r)$ of a point mass $M$ against the distance $r$.

Which of the five graphs is it?

Relevant theory

The potential at a point is the work done in bringing 1 kg from infinity to that point.

At infinity $V = 0$.

Gravity pulls. So the field does the work.

That work is negative. So $V < 0$ at every distance.

The negative sign above indicates that the relevant work is done by the field and not against the field.

Required equations

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

$V(r) = -\int_{\infty}^{r} \frac{GM}{r^2}\,dr$

$V(r) = -\frac{GM}{r}$

Reasoning and calculation

The graph has two axes.

Across: the distance $r$ from the mass.

Up: the potential $V(r)$.

Above the zero line $V$ is positive. Below it $V$ is negative.

Start with a small $r$.

$$V = -\dfrac{GM}{r}$$

When $r$ is small, $V$ is a large negative number.

So the curve starts far below the zero line.

Is there a place where the rule changes?

No. Outside the mass the same rule holds for every $r$.

So it is one smooth curve.

Now make $r$ bigger.

$\dfrac{GM}{r}$ gets smaller. So $V$ stays negative and rises towards zero.

Double $r$ and $|V|$ becomes half.

What happens at the two ends?

When $r \to 0$, $V \to -\infty$.

When $r \to \infty$, $V \to 0$.

But $V$ is not zero at any fixed distance. It only gets closer and closer to zero.

How steep is the curve?

$$\dfrac{dV}{dr} = \dfrac{GM}{r^2}$$

This is always positive. So the curve always goes up.

It is steep when $r$ is small. It is flat when $r$ is large.

Reading the options

Look at the five graphs.

Only graph (2) starts far below zero, rises steeply, then flattens and gets close to zero.

So the answer is graph (2).

Why the other options fail

The shape is right. But $V$ is above zero.

$V$ is positive and reaches zero at a finite $r$.

It is negative. But it meets the axis at a finite $r$.

A constant $V$ means no field.

Conclusion

$$V(r) = -\dfrac{GM}{r}$$

It is always below zero. It rises towards zero as $r$ grows.

So the answer is graph (2).

Drawing the potential as a positive $1/r$ curve.

Gravity always pulls. So the potential is negative.