2023 A/L Physics Past Paper · Question 19 (English Medium)

Answer: (3) $\sqrt{\dfrac{r_2}{r_1}}$

What the question asks

What we know:

The Earth goes round the Sun on a circle of radius $r_1$, with speed $v_1$.

Mars goes round the Sun on a circle of radius $r_2$, with speed $v_2$.

There are no numbers. We keep the letters, and later we check with a sample: a path for Mars that is $4$ times as big as the path of the Earth.

We want: the ratio $\dfrac{v_1}{v_2}$ of the two speeds.

What each letter means:

$G$ ($\text{N m}^2\,\text{kg}^{-2}$): the gravitational constant.

$M$ ($\text{kg}$): the mass of the Sun.

$m$ ($\text{kg}$): the mass of a planet.

$r$ ($\text{m}$): the radius of the path of a planet.

$v$ ($\text{m s}^{-1}$): the speed of a planet.

Relevant theory

A young scientist Isac Newton who lived during the mid era of the same century carrying investigations on the approximate circular motion of the planets in the solar system, around the Sun revealed that a certain force of attraction made by the sun on them helps in their circular motions.

The gravitational force of attraction between two particles of masses M and m situated at a distance r is given by,

This force is thus known as "centripetal force" and is given by. $F = ma$ $$F = \frac{mv^2}{r} = mr\omega^2$$

The quantities G and M are constants. Hence from the above expression it is seen that the speed of a satellite and its period of revolution depend only on the radius of its orbit.

The first quote says the Sun pulls on a planet, and that this pull helps the planet go round. The second quote tells us what a pull between two bodies depends on. The third quote gives the force a body needs to stay on a circle. The last quote says the speed depends on the radius.

The last quote is about a satellite going round the Earth. A planet going round the Sun is the same kind of motion. The Sun takes the place of the Earth, and the pull of the Sun is the force that holds the planet on its circle.

So the pull of the Sun on the planet is equal to the force $\dfrac{mv^2}{r}$ that the circle needs.

Required equations

$F = G\dfrac{Mm}{r^2}$

$F = \dfrac{mv^2}{r}$

$G\dfrac{Mm}{r^2} = \dfrac{mv^2}{r}$

$v^2 = \dfrac{GM}{r}$

Reasoning and calculation

What stays the same: $G$ and the mass $M$ of the Sun. The mass $m$ of the planet has cancelled, so it does not matter.

What changes: the radius $r$ of the path, and with it the speed $v$.

For example, think of toy cars going round a big magnet in the middle of a table, held on their circles by the pull.

A car near the magnet feels a strong pull, so it must go fast to stay on its circle. A car far from the magnet feels a weak pull, so it goes slowly.

A planet near the Sun is like the car near the magnet.

Write the law for each planet, then divide.

$$v_1^2 = \dfrac{GM}{r_1} \qquad v_2^2 = \dfrac{GM}{r_2}$$

$$\dfrac{v_1^2}{v_2^2} = \dfrac{GM/r_1}{GM/r_2} = \dfrac{r_2}{r_1}$$

The $G$ and the $M$ cancel.

This is the ratio of the speeds squared. Take the square root of both sides:

$$\dfrac{v_1}{v_2} = \sqrt{\dfrac{r_2}{r_1}}$$

A check with a sample. Let the path of Mars be $4$ times as big, $r_2 = 4r_1$:

$$\dfrac{v_1}{v_2} = \sqrt{4} = 2$$

The Earth, which is closer, moves $2$ times as fast as Mars.

Reading the options

The ratio is $\dfrac{v_1}{v_2} = \sqrt{\dfrac{r_2}{r_1}}$. That is option (3).

Options (1) and (4) have the ratio of the radii upside down. Options (2) and (5) have the wrong power.

Checks

Dimension check:

$\dfrac{v_1}{v_2}$ is a speed divided by a speed, and $\dfrac{r_2}{r_1}$ is a length divided by a length. Both are plain numbers, so they can be put equal.

$pow(k, 0.5)$

Does it make sense?

The Earth is closer to the Sun than Mars is, so $r_2$ is bigger than $r_1$. Then $\dfrac{r_2}{r_1}$ is more than $1$, and so is its square root. The Earth is the faster one, which fits.

Conclusion

$\dfrac{v_1}{v_2} = \sqrt{\dfrac{r_2}{r_1}}$

$\dfrac{v_1^2}{v_2^2} = \dfrac{r_2}{r_1}$, so $\dfrac{v_1}{v_2} = \sqrt{\dfrac{r_2}{r_1}}$.

The answer is (3).

A body held on a circle by gravity has $v^2 = \dfrac{GM}{r}$. So $v$ goes with $\dfrac{1}{\sqrt{r}}$.

Write the ratio of the two cases, and take the square root. A closer planet must be the faster one.

Exam tip