Physics 9702/42 — May/June 2016

Background Concept

In a binary system, each star is kept in circular motion by a centripetal force directed towards the centre of its circular path.

If the only significant interaction is gravity, the centripetal force is just the gravitational force of attraction between the two masses:

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

Newton’s third law says that forces in an interaction pair are equal in magnitude and opposite in direction: “force on A due to B” equals “force on B due to A” in magnitude.

Understanding the Question

We are told both stars move in circles about the same point P. The question asks why the centripetal forces on A and B are equal in magnitude.

Even though their masses and orbit radii may be different, the force responsible for both circular motions comes from the same two-body interaction.

Approach

1. Identify what provides the centripetal force (gravity between the stars).
2. Use Newton’s third law to compare the force on A with the force on B.

Step-by-Step Reasoning

• Star A attracts star B gravitationally; star B attracts star A gravitationally.
• These two forces form a Newton’s third law pair, so their magnitudes are equal.
• Each force acts towards the other star (along the line joining their centres), which is also the direction towards the centre of the circular motion about P at that instant.
• Therefore, the centripetal force on A and the centripetal force on B have the same magnitude (but act on different bodies, in opposite directions).

Key Takeaways

• In two-body gravity problems, the forces on the two bodies are always equal in magnitude (Newton’s third law).
• The centripetal force is not a new force; it is the name for the resultant inward force causing circular motion.

Common Mistakes

• Saying the centripetal forces are equal because the stars have the same angular speed (that alone does not force equal centripetal force because $F = m r \omega^2$ depends on $m$ and $r$).
• Mixing up “same magnitude” with “same direction” (directions are opposite).

Things to Be Careful About

• Be explicit that the centripetal force is the gravitational attraction.
• State Newton’s third law clearly: equal magnitude, opposite direction, acting on different bodies.

Background Concept

For uniform circular motion, the angular speed $\omega$ is the rate of change of angular displacement (in radians) per second. One complete orbit corresponds to an angular displacement of $2\pi$ radians, so

$$\omega = \frac{2\pi}{T}$$

where $T$ is the period (time for one full orbit).

Understanding the Question

The stars complete one orbit about point P in $4.0$ years. We need the angular speed in $\text{rad s}^{-1}$, so the period must be converted into seconds before using $\omega = 2\pi/T$.

Approach

1. Convert $T = 4.0$ years into seconds.
2. Substitute into $\omega = 2\pi/T$.
3. Round appropriately (typically 2 s.f. because $4.0$ has 2 s.f.).

Step-by-Step Reasoning

• Convert years to seconds:

$$T = 4.0 \times 365\ \text{days} \times 24\ \text{h day}^{-1} \times 3600\ \text{s h}^{-1}$$ $$T = 1.26 \times 10^8\ \text{s}$$

• Now calculate angular speed:

$$\omega = \frac{2\pi}{1.26 \times 10^8} = 4.98 \times 10^{-8}\ \text{rad s}^{-1}$$

• Rounding to 2 significant figures gives:

$$\omega \approx 5.0 \times 10^{-8}\ \text{rad s}^{-1}$$

Key Takeaways

• Always convert the period into SI units (seconds) when angular speed is required in $\text{rad s}^{-1}$.
• One full revolution is $2\pi$ radians.

Common Mistakes

• Leaving the period in years and giving $\omega$ in $\text{rad year}^{-1}$.
• Using $\omega = \pi/T$ instead of $2\pi/T$.
• Rounding too early (carry at least 3 s.f. in intermediate steps).

Things to Be Careful About

• Use standard form correctly when dividing by a large number.
• The unit is $\text{rad s}^{-1}$ (radian is dimensionless but the unit is still written).

Background Concept

If two objects orbit each other under mutual forces, they actually orbit about their common centre of mass. For two masses on a line,

$$M_A r_A = M_B r_B$$

where $r_A$ and $r_B$ are their distances from the centre of mass (here point P), measured along the line joining them.

Also, the total separation of their centres is

$$r_A + r_B = R$$

where $R$ is the distance between the stars.

Understanding the Question

The distance between the star centres is $R = 2.8 \times 10^8\ \text{km}$. Point P is the common centre of the circular orbits (centre of mass). The distance from A to P is $d$, so $r_A = d$ and $r_B = R - d$.

We are given $M_A/M_B = 3.0$ and asked to find $d$.

Approach

1. Use the centre-of-mass condition $M_A d = M_B (R-d)$.
2. Substitute the mass ratio to get a single equation in $d$.
3. Solve for $d$.

Step-by-Step Reasoning

• Write the centre-of-mass relation about P:

$$M_A d = M_B (R - d)$$

• Use $M_A = 3.0 M_B$:

$$(3.0 M_B)d = M_B(R-d)$$

Cancel $M_B$:

$$3d = R - d$$

• Solve:

$$4d = R = 2.8 \times 10^8\ \text{km}$$ $$d = \frac{2.8 \times 10^8}{4} = 7.0 \times 10^7\ \text{km}$$

This makes sense: the more massive star A (three times heavier) is closer to P, so $d$ is smaller than $R/2$.

Key Takeaways

• Two-body circular motion happens about the centre of mass.
• The condition $M_A r_A = M_B r_B$ is extremely useful with given mass ratios.

Common Mistakes

• Using $M_A/M_B = d/(R-d)$ instead of the correct inverted ratio (it should be $M_A/M_B = (R-d)/d$).
• Assuming P is midway between the stars (only true if masses are equal).

Things to Be Careful About

• Keep distances consistent: $r_B = R - d$ (not $R + d$).
• Units: here the question wants $d$ in km, so no conversion is needed in this part.

Background Concept

For circular orbits, the required centripetal force is

$$F_c = m r \omega^2$$

where $m$ is the mass of the orbiting object, $r$ is its orbital radius about the centre of its circular path, and $\omega$ is the angular speed.

In a binary star system, the inward force causing the circular motion is the gravitational attraction between the stars:

$$F_g = \frac{G M_A M_B}{R^2}$$

where $R$ is the separation of their centres.

So we set $F_g = F_c$ for either star, using the correct radius about point P.

Understanding the Question

We know:

• Period $T = 4.0$ years, so from (a)(ii) we have $\omega$.
• Total separation of stars: $R = 2.8 \times 10^8\ \text{km}$.
• From (b)(i), distance from A to P: $d = 7.0 \times 10^7\ \text{km}$.
• Mass ratio: $M_A/M_B = 3.0$.

We must find $M_B$ in kg. This needs SI units: km to m, and use $G = 6.67 \times 10^{-11}\ \text{N m}^2\ \text{kg}^{-2}$.

Approach

1. Find star B’s orbital radius about P: $r_B = R - d$.
2. Write gravitational force between the stars and equate it to the centripetal force for star B.
3. Cancel $M_B$ where possible and use $M_A = 3M_B$ to solve for $M_B$.

Step-by-Step Reasoning

• Convert separation to metres:

$$R = 2.8 \times 10^8\ \text{km} = 2.8 \times 10^{11}\ \text{m}$$

• Find star B’s orbital radius (distance from B to point P):

$$r_B = (2.8 \times 10^8 - 7.0 \times 10^7)\ \text{km} = 2.1 \times 10^8\ \text{km} = 2.1 \times 10^{11}\ \text{m}$$

• Equate forces for star B:

$$\frac{G M_A M_B}{R^2} = M_B r_B \omega^2$$

• Cancel $M_B$ (this is common in orbit problems):

$$\frac{G M_A}{R^2} = r_B \omega^2$$

• Substitute $M_A = 3M_B$:

$$\frac{G (3M_B)}{R^2} = r_B \omega^2$$

• Rearrange for $M_B$:

$$M_B = \frac{r_B \omega^2 R^2}{3G}$$

• Substitute numbers (using $\omega = 4.98 \times 10^{-8}\ \text{rad s}^{-1}$):

$$M_B \approx 2.0 \times 10^{29}\ \text{kg}$$

The result is of the order of a solar mass ($\approx 2 \times 10^{30}\ \text{kg}$), which is reasonable for a star.

Key Takeaways

• For circular gravitational orbits, set gravitational attraction equal to centripetal requirement.
• Use the separation $R$ in the gravitational force, but the orbital radius about the centre of mass ($r_A$ or $r_B$) in the centripetal term.
• Mass ratios allow you to eliminate one mass.

Common Mistakes

• Using $r_B = d$ instead of $r_B = R - d$.
• Mixing up the two different distances: $R$ (between stars) vs. $r_B$ (from star B to P).
• Forgetting to convert km to m, causing the mass to be wrong by a factor of $10^3$ or $10^6$.
• Not cancelling $M_B$ correctly, leading to unnecessary algebra.

Things to Be Careful About

• Keep SI units throughout when using $G$.
• Use the angular speed from (a)(ii) in $\text{rad s}^{-1}$.
• Square brackets and powers: $R^2$ is very large; handle standard form carefully to avoid exponent errors.