Q.Let us assume that our galaxy consists of $2.5 \times 10^{11}$ stars each of one solar mass. How long will a star at a distance of 50,000 ly from the galactic centre take to complete one revolution? Take the diameter of the Milky Way to be $10^{5}$ ly.

Concept and Intuition

For any body in circular orbit around the Earth, gravity supplies the centripetal force, which leads directly to Kepler's third law: $T^2 \propto R^3$, i.e. $T \propto R^{3/2}$. For a small fractional change in radius, the fractional change in period scales by the power (3/2), via the binomial approximation $(1+x)^n \approx 1 + nx$ for small $x$.

Step-by-Step Solution

1. $T \propto R^{3/2}$, so $\dfrac{T_2}{T_1} = \left(\dfrac{R_2}{R_1}\right)^{3/2}$.
2. Here $R_2 = 1.01\,R_1$, so $\dfrac{T_2}{T_1} = (1.01)^{3/2}$.
3. Using the binomial approximation: $(1+0.01)^{1.5} \approx 1 + 1.5 \times 0.01 = 1.015$.
4. So $T_2$ is about 1.5% larger than $T_1$.

Common Mistakes …

Concept and Intuition

The standard orbital period formula involves $GM$ (the planet's gravitational parameter), not $g$ and $R$ directly. But since surface gravity is $g = GM/R^2$, we can substitute $GM = gR^2$ to re-express the period purely in terms of the given variables $a$, $g$, $R$ — this is a dimensional/algebraic substitution trick rather than new physics.

Step-by-Step Solution

1. Start from Kepler's third law (derived from gravity providing centripetal force):

2. Use $g = \dfrac{GM}{R^2}$ (definition of surface gravity), so $GM = gR^2$.
3. Substitute: $T = 2\pi\sqrt{\frac{a^3}{gR^2}} = 2\pi\, a^{3/2}\, (gR^2)^{-1/2} = 2\pi\, a^{3/2}\, g^{-1/2}\, R^{-1}$ …

Concept and Intuition

When a quantity depends on several others through an unknown power law, dimensional analysis pins down the powers by requiring the dimensions on both sides to match exactly, since a physical equation must be dimensionally homogeneous.

Step-by-Step Solution

1. Assume $T = K\,R^a M^b G^c$ (K dimensionless).
2. Dimensions: $[T]=\text{T}$, $[R]=\text{L}$, $[M]=\text{M}$, $[G]=\text{M}^{-1}\text{L}^3\text{T}^{-2}$.
3. Equate: $\text{T}^1\text{M}^0\text{L}^0 = \text{L}^{a}\,\text{M}^{b}\,\left(\text{M}^{-1}\text{L}^{3}\text{T}^{-2}\right)^{c}$.
4. Collect powers: L: $a+3c=0$; M: $b-c=0$; T: $-2c=1$. …

Concept and Intuition

For a satellite in circular orbit, gravity supplies the centripetal force, which directly links its kinetic energy to the orbital radius. Kepler's third law separately links the orbital radius to the period. Chaining the two relations connects kinetic energy to the period.

Step-by-Step Solution

1. Gravity provides centripetal force: $\dfrac{mv^2}{R} = \dfrac{GMm}{R^2} \Rightarrow v^2 = \dfrac{GM}{R}$.
2. Kinetic energy: $KE = \tfrac12mv^2 = \dfrac{GMm}{2R} \propto \dfrac1R$.
3. Kepler's third law: $T^2\propto R^3 \Rightarrow R\propto T^{2/3}$. …

Concept and Intuition

For a satellite in a circular orbit, gravitational attraction is exactly the centripetal force needed to keep it in that orbit: $\dfrac{GMm}{r^2}=\dfrac{mv^2}{r}$, which gives the orbital speed $v=\sqrt{GM/r}$ — notably independent of the satellite's own mass.

Step-by-Step Solution

1. $GM = (6.67\times10^{-11})(6\times10^{24}) = 4.002\times10^{14}\ \mathrm{m^3 s^{-2}}$.
2. $v=\sqrt{GM/r}=\sqrt{\dfrac{4.002\times10^{14}}{6.67\times10^{6}}}$.
3. $\dfrac{4.002\times10^{14}}{6.67\times10^{6}}\approx6.0\times10^{7}$. …

Concept and Intuition

For a satellite in circular orbit, gravity supplies the centripetal force: $\dfrac{GMm}{r^2}=\dfrac{mv^2}{r}=m\omega^2 r$, leading to $T=2\pi\sqrt{r^3/GM}$ where $r$ is the orbital radius measured from Earth's centre (so altitude must be added to Earth's radius).

Step-by-Step Solution

1. $r = R_e + h = 6.4\times10^6 + 1\times10^6 = 7.4\times10^6$ m.
2. $r^3 = (7.4\times10^6)^3 = 405.224\times10^{18} = 4.05224\times10^{20}\ \mathrm{m^3}$.
3. $GM = 6.67\times10^{-11}\times6\times10^{24} = 4.002\times10^{14}\ \mathrm{m^3s^{-2}}$. …

Concept and Intuition

For any central force law $F = k/R^n$ providing the centripetal force for a circular orbit, we can derive how the orbital period scales with radius by equating $F$ to $m\omega^2 R$ and solving for $\omega$, then converting to $T = 2\pi/\omega$.

Step-by-Step Solution

1. Required centripetal force for circular motion: $F = m\omega^2 R$.
2. Given force law: $F = \dfrac{k}{R^3}$ (inverse cube).
3. Equate: $m\omega^2 R = \dfrac{k}{R^3} \Rightarrow \omega^2 = \dfrac{k}{mR^4} \Rightarrow \omega \propto R^{-2}$. …

Concept and Intuition

For circular orbits around a common star, Kepler's third law gives $T^2\propto r^3$. Orbital speed is $v=\dfrac{2\pi r}{T}$. Eliminating $r$ in favor of $T$ lets us directly compare speeds using only the period ratio.

Step-by-Step Solution

1. Kepler's third law: $T^2 \propto r^3 \Rightarrow r \propto T^{2/3}$.
2. Orbital speed: $v = \dfrac{2\pi r}{T} \propto \dfrac{T^{2/3}}{T} = T^{-1/3}$.
3. Given $T_A = 8T_B$: $\dfrac{v_A}{v_B} = \left(\dfrac{T_A}{T_B}\right)^{-1/3} = 8^{-1/3} = \dfrac12$. …

Concept and Intuition

A satellite in a geostationary orbit orbits in the equatorial plane, in the

same direction as Earth's rotation, at exactly the altitude (about 35,786 km)

where Kepler's third law gives an orbital period equal to Earth's sidereal

rotation period. Because the satellite's angular position tracks Earth's

rotation exactly, it appears to hang motionless over one longitude — which is

precisely why it must have a period of 24 hours (approximating the sidereal

day as 24 h for this level of question).

Step-by-Step Solution

1. "Geostationary" is defined operationally: the satellite must remain above the same point on Earth's surface at all times.
2. This is only possible if the satellite's angular velocity equals Earth's angular velocity of rotation.
3. Earth completes one rotation (one day) in 24 hours. …

Concept and Intuition

A geostationary satellite orbits Earth with a period equal to Earth's rotation period, $T=24$ hours, at an orbital radius corresponding to a height of about $7R$ above the surface — i.e., orbital radius $r_1 = R + 7R = 8R$ from Earth's centre.

Kepler's third law states that for objects orbiting the same central body (Earth here), $T^2 \propto r^3$, i.e., $\dfrac{T_2^2}{T_1^2} = \left(\dfrac{r_2}{r_1}\right)^3$. This lets us find the new period once the new orbital radius is known, without needing to know $G$ or $M$ explicitly.

Step-by-Step Solution

1. Original orbital radius (geostationary): $r_1 = R + 7R = 8R$, with $T_1 = 24$ h.
2. New orbital radius (height $R$ above surface): $r_2 = R + R = 2R$. …