Dynamics and Space

Higher Dynamics

Explore the simulation above to develop intuition for this topic.

Motion in a Straight Line

Displacement, velocity, and acceleration are related through the kinematic equations (covered in Mechanics).

Graphical Analysis

Displacement-Time Graph:

• Gradient = velocity
• Curved line = changing velocity (acceleration)

Velocity-Time Graph:

• Gradient = acceleration
• Area under the graph = displacement

Example: A car accelerates from rest at 2 \mathrm{ m/s^2 for 5 \mathrm{ sThen travels at Constant speed for 10 \mathrm{ sThen decelerates at 3 \mathrm{ m/s^2 to rest. Find the total Distance.

Phase 1: s_1 = \frac{1}{2}(0 + 10)(5) = 25 \mathrm{ m

Phase 2: s_2 = 10 \times 10 = 100 \mathrm{ m

Phase 3: t_3 = \dfrac{0 - 10}{-3} = \dfrac{10}{3} \mathrm{ s s_3 = \frac{1}{2}(10 + 0)\left(\frac{10}{3}\right) = \frac{50}{3} \approx 16.67 \mathrm{ m

Total: 25 + 100 + 16.67 = 141.67 \mathrm{ m

Extracting Information from Graphs

The gradient of a tangent to a curved displacement-time graph gives the instantaneous velocity at That point. The area under a velocity-time graph (even a curved one) gives the total displacement. For non-constant acceleration, count squares or use integration.

Circular Motion

For an object moving in a circle of radius $r$ at speed $v$:

Angular velocity: $\omega = \dfrac{v}{r} = \dfrac{2\pi}{T} = 2\pi f$

Centripetal acceleration: $a = \dfrac{v^2}{r} = \omega^2 r$

Centripetal force: $F = \dfrac{mv^2}{r} = m\omega^2 r$

This force always acts towards the centre of the circle.

Why the Centripetal Force Points Inward

Newton’s first law says an object continues in a straight line unless acted upon by a force. In Circular motion, the object is continuously deflected from its straight-line path toward the centre. The force causing this deflection is the centripetal force. It is not a separate interaction; it is The name for whatever force or combination of forces provides the net inward force (gravity for Orbits, friction for a car on a roundabout, tension for a ball on a string).

Why Objects Released from Circular Motion Move Tangentially

When the centripetal force is removed (the string breaks, friction is insufficient), the object Continues in a straight line tangent to the circle at the point of release. This is Newton’s first Law: with no net force, the object continues with constant velocity. The velocity at the moment of Release is tangential to the circle, so the object follows a tangent.

Example: A car of mass 1200 \mathrm{ kg travels around a roundabout of radius 25 \mathrm{ m at 12 \mathrm{ m/s. Find the centripetal force.

F = \frac{mv^2}{r} = \frac{1200 \times 144}{25} = \frac{172800}{25} = 6912 \mathrm{ N

This force is provided by friction between the tyres and the road.

Vertical Circular Motion

At the top of a vertical circle, both weight and tension (or normal force) point toward the centre:

$F_c = mg + T = \frac{mv^2}{r}$

At the bottom, tension points toward the centre but weight points away:

$F_c = T - mg = \frac{mv^2}{r}$

The minimum speed at the top to maintain contact (when $T = 0$):

$v_{\min} = \sqrt{gr}$

Orbital Motion and Satellites

Example: A satellite orbits Earth at an altitude of 400 \mathrm{ km in a circular orbit. Given Earth’s mass M = 5.97 \times 10^{24} \mathrm{ kg and radius R = 6.37 \times 10^6 \mathrm{ mFind The orbital speed and period.

r = R + h = 6.37 \times 10^6 + 4 \times 10^5 = 6.77 \times 10^6 \mathrm{ m

$\frac{GMm}{r^2} = \frac{mv^2}{r}$

$v = \sqrt{\frac{GM}{r}} = \sqrt{\frac{6.67 \times 10^{-11} \times 5.97 \times 10^{24}}{6.77 \times 10^6}}$

v = \sqrt{\frac{3.983 \times 10^{14}}{6.77 \times 10^6}} = \sqrt{5.883 \times 10^7} \approx 7670 \mathrm{ m/s

T = \frac{2\pi r}{v} = \frac{2\pi \times 6.77 \times 10^6}{7670} \approx 5544 \mathrm{ s \approx 92.4 \mathrm{ min

Gravitational Fields

Newton’s Law of Gravitation:

$F = \frac{Gm_1 m_2}{r^2}$

Where G = 6.67 \times 10^{-11} \mathrm{ N m^2 \mathrm{kg^{-2}.

Why Gravity Is an Inverse-Square Law

Consider a point mass emitting gravitational field lines uniformly in all directions. The number of Field lines crossing a sphere of radius $r$ is constant (the same lines cross every sphere centered On the mass). The flux density (field lines per unit area) is proportional to $1/r^2$Since the Surface area of the sphere is $4\pi r^2$. Therefore, the gravitational field strength is Proportional to $1/r^2$.

Gravitational Field Strength:

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

At Earth’s surface: g \approx 9.8 \mathrm{ N/kg.

Gravitational Potential Energy:

$E_p = -\frac{GMm}{r}$

This is zero at infinity and negative for finite distances. The negative sign means energy must be Supplied to move masses apart.

Why Gravitational PE Is Negative

The convention $U = 0$ at $r = \infty$ is natural because the gravitational force decreases to zero At infinity. As two masses are brought together from infinity, gravity does positive work and the Potential energy decreases below zero. A bound orbit (like a planet around a star) has negative Total energy.

Escape Velocity:

v_{\mathrm{esc} = \sqrt{\frac{2GM}{r}}

Example: Calculate the escape velocity from Earth’s surface.

v_{\mathrm{esc} = \sqrt{\frac{2 \times 6.67 \times 10^{-11} \times 5.97 \times 10^{24}}{6.37 \times 10^6}} = \sqrt{\frac{7.966 \times 10^{14}}{6.37 \times 10^6}} = \sqrt{1.251 \times 10^8} \approx 11180 \mathrm{ m/s \approx 11.2 \mathrm{ km/s

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Higher Astrophysics

Stellar Evolution

Stars are born from clouds of gas and dust (nebulae) that collapse under gravity.

Main Sequence: A star is in equilibrium when the outward radiation pressure balances Gravitational collapse. Hydrogen fuses to form helium in the core.

4^1_1\mathrm{H \to ^4_2\mathrm{He + 2^0_{+1}\mathrm{e + 2\nu_e + \mathrm{energy

Evolution depends on mass:

• Low-mass stars (like the Sun): Main sequence $\to$ Red giant $\to$ Planetary nebula $\to$ White dwarf
• High-mass stars ($> 8$ solar masses): Main sequence $\to$ Red supergiant $\to$ Supernova $\to$ Neutron star or Black hole

Why Massive Stars Have Shorter Lives

A more massive star has a higher core temperature and pressure, so it burns its fuel much faster. Although it has more fuel, the rate of consumption increases more than proportionally with mass. A Star of 10 solar masses may live only 20 million years, compared to 10 billion years for the Sun. The luminosity of a main-sequence star scales roughly as $L \propto M^{3.5}$While the available Fuel scales as $M$So the lifetime scales as $M^{-2.5}$.

The Hertzsprung-Russell Diagram

The HR diagram plots luminosity (or absolute magnitude) against temperature (or spectral class).

• Main sequence: Diagonal band from top-left (hot, bright) to bottom-right (cool, dim)
• Red giants: Top-right (cool but bright, large radius)
• White dwarfs: Bottom-left (hot but dim, small radius)

Wien’s Law and Stefan-Boltzmann Law

Wien’s Displacement Law:

\lambda_{\mathrm{max} T = 2.898 \times 10^{-3} \mathrm{ m K

Example: A star has surface temperature 5800 \mathrm{ K. Find the peak wavelength of its Radiation.

\lambda_{\mathrm{max} = \frac{2.898 \times 10^{-3}}{5800} = 5.0 \times 10^{-7} \mathrm{ m = 500 \mathrm{ nm

This is in the visible spectrum (green-yellow).

Stefan-Boltzmann Law:

$L = 4\pi r^2 \sigma T^4$

Where \sigma = 5.67 \times 10^{-8} \mathrm{ W m^{-2} \mathrm{K^{-4}.

Universal Gravitation and Orbits

Kepler’s Third Law:

$T^2 \propto r^3$

$T^2 = \frac{4\pi^2}{GM}r^3$

Example: A planet orbits a star of mass 2 \times 10^{30} \mathrm{ kg at a distance of 1.5 \times 10^{11} \mathrm{ m. Find the orbital period.

$T = 2\pi\sqrt{\frac{r^3}{GM}} = 2\pi\sqrt{\frac{(1.5 \times 10^{11})^3}{6.67 \times 10^{-11} \times 2 \times 10^{30}}}$

$= 2\pi\sqrt{\frac{3.375 \times 10^{33}}{1.334 \times 10^{20}}} = 2\pi\sqrt{2.530 \times 10^{13}}$

= 2\pi \times 5.030 \times 10^6 \approx 3.16 \times 10^7 \mathrm{ s \approx 365.7 \mathrm{ days

Cosmology

Hubble’s Law:

$v = H_0 d$

Where H_0 \approx 70 \mathrm{ km s^{-1} \mathrm{Mpc^{-1} is the Hubble constant, $v$ is the Recession velocity, and $d$ is the distance.

Redshift: As the universe expands, light from distant galaxies is shifted to longer wavelengths.

z = \frac{\Delta\lambda}{\lambda} = \frac{v}{c} \quad (\mathrm{for v \ll c)

Age of the Universe:

t \approx \frac{1}{H_0} \approx 14 \mathrm{ billion years

Big Bang Theory:

• The universe began from a singularity approximately 13.8 billion years ago
• Evidence: cosmic microwave background radiation, Hubble’s law, abundance of light elements
• The universe has been expanding and cooling ever since

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Common Pitfalls

1. Centripetal force is not a separate force: It is the resultant force directed towards the centre (e.g., tension, friction, gravity).

2. Gravitational force vs. Gravitational field strength: $F = \dfrac{GMm}{r^2}$ (force on mass $m$) vs. $g = \dfrac{GM}{r^2}$ (field strength, force per unit mass).

3. Sign of gravitational PE: Gravitational PE is negative ($E_p = -\dfrac{GMm}{r}$) because work must be done to move objects apart.

4. Hubble’s law units: Ensure consistent units when using $v = H_0 d$.

5. Stellar evolution paths: Low-mass and high-mass stars follow different evolutionary paths after the main sequence.

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Practice Questions

1. A 1500 \mathrm{ kg car travels around a bend of radius 50 \mathrm{ m at 15 \mathrm{ m/s. Find the centripetal force and the minimum coefficient of friction needed.

2. Find the gravitational field strength at a height of 300 \mathrm{ km above Earth’s surface.

3. A star has peak wavelength 350 \mathrm{ nm. Find its surface temperature and state what colour it appears.

4. Calculate the escape velocity from Mars (mass 6.39 \times 10^{23} \mathrm{ kgRadius 3.39 \times 10^6 \mathrm{ m).

5. A galaxy is observed to have a redshift $z = 0.02$. Find its recession velocity and approximate distance.

6. Explain why a white dwarf is hot but has low luminosity.

7. A satellite orbits at a height of 500 \mathrm{ km above Earth. Find its orbital speed and period.

8. Compare and contrast the evolution of a star of 1 solar mass with a star of 15 solar masses.

9. A ball of mass 0.5 \mathrm{ kg is attached to a string of length 1.0 \mathrm{ m and swung in a vertical circle. Find the minimum speed at the top of the circle, and the tension in the string at the bottom if the speed is 6 \mathrm{ m/s there.

10. A geostationary satellite orbits at a height of 35,786 \mathrm{ km above Earth’s equator. Calculate its orbital speed and period, and explain why it remains above the same point on Earth’s surface.

11. Worked Example: Non-Uniform Circular Motion

A ball of mass 0.5 \mathrm{ kg on a string of length 1.0 \mathrm{ m is swung in a vertical circle. Its speed at the bottom is 8 \mathrm{ m/s. Find the tension in the string at the bottom and the Minimum speed at the top.

At the bottom: The tension and weight act in opposite directions radially.

$T - mg = \frac{mv^2}{r}$

T = mg + \frac{mv^2}{r} = 0.5 \times 9.8 + \frac{0.5 \times 64}{1.0} = 4.9 + 32 = 36.9 \mathrm{ N

At the top: Energy conservation from bottom to top (height difference = 2r = 2 \mathrm{ m):

$\frac{1}{2}mv_b^2 = \frac{1}{2}mv_t^2 + mg(2r)$

$v_t^2 = v_b^2 - 4gr = 64 - 4 \times 9.8 \times 1.0 = 64 - 39.2 = 24.8$

v_t = 4.98 \mathrm{ m/s

Minimum speed at the top: Set $T = 0$ at the top:

\frac{mv_{\min}^2}{r} = mg \implies v_{\min} = \sqrt{gr} = \sqrt{9.8} = 3.13 \mathrm{ m/s

Since v_t = 4.98 \mathrm{ m/s \gt 3.13 \mathrm{ m/s = v_{\min}The string remains taut at the top.

12. Gravitational Fields: Extended Analysis

Worked Example: Gravitational Field Strength at Altitude

Calculate $g$ at an altitude of 300 \mathrm{ km above Earth’s surface.

r = R_E + h = 6.37 \times 10^6 + 3 \times 10^5 = 6.67 \times 10^6 \mathrm{ m

g = \frac{GM}{r^2} = \frac{6.67 \times 10^{-11} \times 5.97 \times 10^{24}}{(6.67 \times 10^6)^2} = \frac{3.983 \times 10^{14}}{4.449 \times 10^{13}} = 8.95 \mathrm{ N/kg

This is about 91% of the surface value of 9.8 \mathrm{ N/kg. The International Space Station, at About 400 km altitude, experiences g \approx 8.7 \mathrm{ N/kg — astronauts are not truly Weightless because there is no gravity at that altitude; they are weightless because they are in Free fall.

Gravitational Potential Energy: Worked Example

A 1000 \mathrm{ kg satellite is moved from an orbit of radius 7000 \mathrm{ km to an orbit of Radius 42000 \mathrm{ km (geostationary). Calculate the energy required.

$\Delta E = E_{p,f} - E_{p,i} = -\frac{GMm}{r_f} - \left(-\frac{GMm}{r_i}\right) = GMm\left(\frac{1}{r_i} - \frac{1}{r_f}\right)$

$= 6.67 \times 10^{-11} \times 5.97 \times 10^{24} \times 1000 \times \left(\frac{1}{7 \times 10^6} - \frac{1}{4.2 \times 10^7}\right)$

$= 3.983 \times 10^{17} \times (1.429 \times 10^{-7} - 2.381 \times 10^{-8})$

= 3.983 \times 10^{17} \times 1.191 \times 10^{-7} = 4.74 \times 10^{10} \mathrm{ J = 47.4 \mathrm{ GJ

This enormous energy requirement explains why launching satellites is so expensive and why ion Thrusters (which provide continuous low thrust) are preferred for orbit adjustments.

13. Stellar Evolution: Extended Analysis

Worked Example: Wien’s Law and Stefan-Boltzmann Law

A star has a radius of 5 \times 10^9 \mathrm{ m (about 7 times the Sun’s radius) and surface Temperature 4000 \mathrm{ K.

Peak wavelength:

\lambda_{\max} = \frac{2.898 \times 10^{-3}}{4000} = 7.25 \times 10^{-7} \mathrm{ m = 725 \mathrm{ nm

This is in the red/infrared part of the spectrum. The star would appear reddish.

Luminosity:

$L = 4\pi r^2 \sigma T^4 = 4\pi(5 \times 10^9)^2 \times 5.67 \times 10^{-8} \times (4000)^4$

$= 4\pi \times 2.5 \times 10^{19} \times 5.67 \times 10^{-8} \times 2.56 \times 10^{14}$

= 4\pi \times 2.5 \times 10^{19} \times 1.452 \times 10^{7} = 4\pi \times 3.63 \times 10^{26} = 4.56 \times 10^{27} \mathrm{ W

This is about 12 times the Sun’s luminosity (3.85 \times 10^{26} \mathrm{ W), consistent with a red Giant star.

The Main Sequence: Mass-Luminosity Relation

For main-sequence stars:

$L \propto M^{3.5}$

A star twice as massive as the Sun is $2^{3.5} \approx 11.3$ times as luminous. Since the available Fuel is proportional to $M$ but the consumption rate is proportional to $L \propto M^{3.5}$The Lifetime is:

\mathrm{lifetime \propto \frac{M}{L} \propto \frac{M}{M^{3.5}} = M^{-2.5}

A star of 10 solar masses lives $10^{-2.5} \approx 0.00316$ times as long as the Sun, or about 32 Million years (compared to 10 billion years for the Sun). This is why massive stars are rare: they Burn through their fuel very quickly.

14. Cosmology: Extended Analysis

Worked Example: Hubble’s Law and Redshift

A galaxy has a redshift $z = 0.05$. Find its recession velocity and distance.

For $z \lt 0.1$: v \approx cz = 3 \times 10^8 \times 0.05 = 1.5 \times 10^7 \mathrm{ m/s = 15000 \mathrm{ km/s

d = \frac{v}{H_0} = \frac{15000 \mathrm{ km/s}{70 \mathrm{ km/s/Mpc} = 214 \mathrm{ Mpc

This galaxy is about 214 megaparsecs away, or roughly 700 million light-years.

The Cosmic Microwave Background

The cosmic microwave background (CMB) radiation was discovered by Penzias and Wilson in 1965. It is Thermal radiation with a black-body temperature of 2.725 \mathrm{ KPeaking in the microwave Region. The CMB is the afterglow of the Big Bang: after the universe became transparent about 380,000 years after the Big Bang, the thermal radiation from the hot, dense early universe has been Redshifted by the expansion of space to microwave wavelengths.

15. Summary Table: Key Formulas

Topic	Formula	Variables	Notes
Centripetal acceleration	$a_c = v^2/r = \omega^2 r$	$v$, $r$, $\omega$	Always toward centre
Centripetal force	$F_c = mv^2/r$	$m$, $v$, $r$	Net inward force
Universal gravitation	$F = GMm/r^2$	$G$, $M$, $m$, $r$	Inverse-square law
Gravitational field	$g = GM/r^2$	$G$, $M$, $r$	Force per unit mass
Gravitational PE	$E_p = -GMm/r$	$G$, $M$, $m$, $r$	Zero at infinity
Orbital speed	$v = \sqrt{GM/r}$	$G$, $M$, $r$	Circular orbit
Kepler’s third law	$T^2 = 4\pi^2 r^3 / (GM)$	$T$, $r$, $G$, $M$	$T^2 \propto r^3$
Escape velocity	$v_e = \sqrt{2GM/r}$	$G$, $M$, $r$	Independent of mass
Wien’s law	$\lambda_{\max}T = 2.898 \times 10^{-3}$	$\lambda_{\max}$, $T$	Peak wavelength
Stefan-Boltzmann	$L = 4\pi r^2\sigma T^4$	L$$r$$\sigma$$T	Total power output
Hubble’s law	$v = H_0 d$	v$$H_0$$d	Recession velocity

16. Practice Questions (Additional)

11. A 2000 \mathrm{ kg satellite in circular orbit at 500 \mathrm{ km altitude needs to move to a higher orbit at 1000 \mathrm{ km altitude. Calculate the energy required using the gravitational potential energy formula.

12. A star has apparent brightness $B$ and is at distance $d$. If the same star were moved to distance $3d$By what factor would its apparent brightness change?

13. Calculate the escape velocity from the surface of the Sun. (Mass of Sun = 1.989 \times 10^{30} \mathrm{ kgRadius of Sun = 6.96 \times 10^8 \mathrm{ m.)

14. A ball of mass 0.3 \mathrm{ kg on a string of length 0.9 \mathrm{ m is swung in a vertical circle with speed 7 \mathrm{ m/s at the bottom. Calculate the tension at the bottom and the speed at the top. Does the string go slack?

15. Explain why a white dwarf is hot but has low luminosity, while a red giant is cool but has high luminosity. Reference the Stefan-Boltzmann law in your answer.

16. A galaxy at a distance of 100 \mathrm{ Mpc has a recession velocity of 7000 \mathrm{ km/s. Calculate the value of the Hubble constant from this single data point.

17. Calculate the gravitational field strength at the surface of Mars and compare it to Earth’s surface gravity. (Mars mass = 6.39 \times 10^{23} \mathrm{ kgMars radius = 3.39 \times 10^6 \mathrm{ m.)

18. Two stars in a binary system orbit their common centre of mass. Star A has mass 2 \times 10^{30} \mathrm{ kg and Star B has mass 1 \times 10^{30} \mathrm{ kg. They are separated by 1.5 \times 10^{11} \mathrm{ m. Calculate the orbital period of the system.

19. Explain the evidence for the Big Bang theory, including the cosmic microwave background radiation and Hubble’s law.

20. A geostationary satellite must have an orbital period of 24 hours. Calculate the required orbital radius and verify that it is approximately 42200 \mathrm{ km from Earth’s centre.

Extended Worked Examples

Example 21: Orbital Transfer (Hohmann Transfer)

A spacecraft of mass 1000 \mathrm{ kg is in a circular low Earth orbit of radius r_1 = 6.67 \times 10^6 \mathrm{ m. It needs to transfer to a circular orbit of radius r_2 = 6.67 \times 10^7 \mathrm{ m. Calculate the total delta-v required.

Step 1: Speed in initial circular orbit

v_1 = \sqrt{\frac{GM}{r_1}} = \sqrt{\frac{6.67 \times 10^{-11} \times 5.97 \times 10^{24}}{6.67 \times 10^6}} = \sqrt{5.97 \times 10^7} \approx 7730 \mathrm{ m/s

Step 2: Speed at perigee of transfer ellipse

The transfer ellipse has semi-major axis a = \frac{r_1 + r_2}{2} = \frac{6.67 \times 10^6 + 6.67 \times 10^7}{2} = 3.669 \times 10^7 \mathrm{ m.

Using the vis-viva equation at perigee ($r = r_1$):

$v_p = \sqrt{GM \left( \frac{2}{r_1} - \frac{1}{a} \right)} = \sqrt{6.67 \times 10^{-11} \times 5.97 \times 10^{24} \times \left( \frac{2}{6.67 \times 10^6} - \frac{1}{3.669 \times 10^7} \right)}$

v_p = \sqrt{3.98 \times 10^{14} \times (2.998 \times 10^{-7} - 2.726 \times 10^{-8})} = \sqrt{3.98 \times 10^{14} \times 2.726 \times 10^{-7}} \approx 10420 \mathrm{ m/s

Step 3: Speed at apogee of transfer ellipse

$v_a = \sqrt{GM \left( \frac{2}{r_2} - \frac{1}{a} \right)} = \sqrt{3.98 \times 10^{14} \times \left( \frac{2}{6.67 \times 10^7} - \frac{1}{3.669 \times 10^7} \right)}$

v_a = \sqrt{3.98 \times 10^{14} \times (2.998 \times 10^{-8} - 2.726 \times 10^{-8})} = \sqrt{3.98 \times 10^{14} \times 2.726 \times 10^{-9}} \approx 1042 \mathrm{ m/s

Step 4: Speed in final circular orbit

v_2 = \sqrt{\frac{GM}{r_2}} = \sqrt{\frac{3.98 \times 10^{14}}{6.67 \times 10^7}} = \sqrt{5.97 \times 10^6} \approx 2443 \mathrm{ m/s

Step 5: Total delta-v

\Delta v_1 = v_p - v_1 = 10420 - 7730 = 2690 \mathrm{ m/s

\Delta v_2 = v_2 - v_a = 2443 - 1042 = 1401 \mathrm{ m/s

\Delta v_{\mathrm{total} = 2690 + 1401 = 4091 \mathrm{ m/s \approx 4.1 \mathrm{ km/s

:::info Hohmann transfers are the most fuel-efficient two-impulse transfer between coplanar circular Orbits. In practice, mission designers often use three-impulse bi-elliptic transfers when the ratio $r_2/r_1$ is very large (greater than about 11.94). :::

Example 22: Stellar Parallax and Distance

A star is observed to have an annual parallax of 0.077 \mathrm{ arcseconds. Calculate the distance To the star in parsecs and light-years.

Step 1: Use the parallax formula

d = \frac{1}{p} \mathrm{ parsecs

Where $p$ is the parallax angle in arcseconds.

d = \frac{1}{0.077} = 12.99 \mathrm{ pc \approx 13.0 \mathrm{ pc

Step 2: Convert to light-years

1 \mathrm{ pc = 3.26 \mathrm{ ly

d = 12.99 \times 3.26 = 42.3 \mathrm{ ly

Step 3: Convert to metres

1 \mathrm{ pc = 3.086 \times 10^{16} \mathrm{ m

d = 12.99 \times 3.086 \times 10^{16} = 4.01 \times 10^{17} \mathrm{ m

:::caution The Gaia space telescope can measure parallaxes down to about 10 \mathrm{ \mu\mathrm{as (microarcseconds), corresponding to distances of about 100 \mathrm{ kpc. Ground-based observations Are limited to parallaxes above about 0.01 \mathrm{ arcseconds. :::

Example 23: Gravitational Potential Energy in Orbit Changes

A satellite of mass 500 \mathrm{ kg moves from a circular orbit of radius r_1 = 7.0 \times 10^6 \mathrm{ m to a circular orbit of radius r_2 = 7.5 \times 10^6 \mathrm{ m. Calculate the energy that must be supplied by the rocket engines.

Step 1: Total energy in each orbit

$E = -\frac{GMm}{2r}$

E_1 = -\frac{6.67 \times 10^{-11} \times 5.97 \times 10^{24} \times 500}{2 \times 7.0 \times 10^6} = -\frac{1.99 \times 10^{17}}{1.4 \times 10^7} = -1.421 \times 10^{10} \mathrm{ J

E_2 = -\frac{1.99 \times 10^{17}}{1.5 \times 10^7} = -1.327 \times 10^{10} \mathrm{ J

Step 2: Energy change

\Delta E = E_2 - E_1 = -1.327 \times 10^{10} - (-1.421 \times 10^{10}) = 9.4 \times 10^8 \mathrm{ J

The total energy becomes less negative (increases), so energy must be supplied: \Delta E = 9.4 \times 10^8 \mathrm{ J.

:::info Counter-, when a satellite fires its engines forward to speed up, it moves to a Higher orbit where it actually travels slower. The kinetic energy decreases, but the gravitational Potential energy increases by more than the kinetic energy decreases. :::

Common Pitfalls Extended

Pitfall 6: Confusing Gravitational Force with Gravitational Field Strength

Gravitational force $F = \frac{GMm}{r^2}$ acts on a specific mass $m$. Gravitational field strength $g = \frac{GM}{r^2}$ is the force per unit mass and does not depend on the test mass. They have The same numerical value at the same location only when the test mass is 1 \mathrm{ kg.

Pitfall 7: Using the Wrong Radius in Orbital Calculations

The radius $r$ in orbital formulas is the distance from the centre of the planet, not the altitude Above the surface. Always add the planet’s radius to the altitude:

r = R_{\mathrm{planet} + h

Pitfall 8: Assuming Kepler’s Third Law Applies in All Situations

Kepler’s third law $T^2 \propto r^3$ applies specifically to objects orbiting the same central Body. For two different central bodies (e.g., a satellite around Earth vs around Mars), the constant Of proportionality changes:

$T^2 = \frac{4\pi^2}{GM} r^3$

Additional Practice Problems

21. A probe is launched from Earth’s surface with escape velocity. Calculate the speed of the probe when it is at a distance of $3R_E$ from Earth’s centre.

22. Two galaxies are observed to be moving apart at 1200 \mathrm{ km/s. Using Hubble’s constant H_0 = 70 \mathrm{ km/s/MpcCalculate the distance to these galaxies in Mpc and in light-years.

23. A 2000 \mathrm{ kg spacecraft in circular orbit at 300 \mathrm{ km altitude fires retro-rockets to deorbit. Calculate the orbital speed before and after a delta-v of -100 \mathrm{ m/sAnd explain qualitatively what happens to the orbit.

24. The Sun has an apparent brightness of 1.37 \times 10^3 \mathrm{ W/m^2 at Earth. Calculate the luminosity of the Sun and its absolute magnitude (reference: M_{\mathrm{ref} = 4.83).

25. Explain how astronomers distinguish between redshift due to the Doppler effect (peculiar velocity) and cosmological redshift due to the expansion of space.

Further Worked Examples

Example 26: Binary Star System

Two stars in a binary system orbit their common centre of mass with a period of 5.0 \mathrm{ years. The semi-major axes of their orbits are 1.5 \times 10^{11} \mathrm{ m and 4.5 \times 10^{11} \mathrm{ m. Calculate the masses of both stars.

Step 1: Use Kepler’s third law for binary systems

$T^2 = \frac{4\pi^2 a^3}{G(m_1 + m_2)}$

Where a = a_1 + a_2 = 1.5 \times 10^{11} + 4.5 \times 10^{11} = 6.0 \times 10^{11} \mathrm{ m.

T = 5.0 \mathrm{ years = 5.0 \times 3.156 \times 10^7 = 1.578 \times 10^8 \mathrm{ s

Step 2: Solve for $m_1 + m_2$

$m_1 + m_2 = \frac{4\pi^2 a^3}{GT^2} = \frac{4\pi^2 \times (6.0 \times 10^{11})^3}{6.67 \times 10^{-11} \times (1.578 \times 10^8)^2}$

= \frac{4 \times 9.8696 \times 2.16 \times 10^{34}}{6.67 \times 10^{-11} \times 2.490 \times 10^{16}} = \frac{8.527 \times 10^{35}}{1.661 \times 10^6} = 5.13 \times 10^{29} \mathrm{ kg

Step 3: Find individual masses (centre of mass)

$m_1 a_1 = m_2 a_2$

$\frac{m_1}{m_2} = \frac{a_2}{a_1} = \frac{4.5 \times 10^{11}}{1.5 \times 10^{11}} = 3$

So $m_1 = 3m_2$.

3m_2 + m_2 = 5.13 \times 10^{29} \implies m_2 = 1.28 \times 10^{29} \mathrm{ kg

m_1 = 3.85 \times 10^{29} \mathrm{ kg

Example 27: Apparent Magnitude and Distance

A star has an apparent magnitude of $6.0$ and an absolute magnitude of $2.0$. Calculate the distance To the star in parsecs and light-years.

Step 1: Use the distance modulus

$m - M = 5\log_{10}\left(\frac{d}{10}\right)$

$6.0 - 2.0 = 5\log_{10}\left(\frac{d}{10}\right)$

$4 = 5\log_{10}\left(\frac{d}{10}\right)$

$\log_{10}\left(\frac{d}{10}\right) = 0.8$

$\frac{d}{10} = 10^{0.8} = 6.31$

d = 63.1 \mathrm{ pc

Step 2: Convert to light-years

d = 63.1 \times 3.26 = 205.7 \mathrm{ ly

:::info The apparent magnitude scale is logarithmic and inverted: a difference of 5 magnitudes Corresponds to a brightness ratio of exactly 100. Each magnitude step represents a factor of $100^{1/5} \approx 2.512$ in brightness. A magnitude 1 star is $2.512^5 = 100$ times brighter than a Magnitude 6 star. :::

Example 28: Gravitational Field Strength Inside a Planet

Assuming Earth has uniform density, calculate the gravitational field strength at a depth of 2000 \mathrm{ km below the surface.

Step 1: Depth below surface

At depth $d$The distance from the centre is r = R_E - d = 6371 - 2000 = 4371 \mathrm{ km.

Step 2: Mass enclosed within radius $r$

For uniform density: M_{\mathrm{enc} = M_E \times \left(\frac{r}{R_E}\right)^3

M_{\mathrm{enc} = 5.97 \times 10^{24} \times \left(\frac{4371}{6371}\right)^3 = 5.97 \times 10^{24} \times (0.686)^3 = 5.97 \times 10^{24} \times 0.323 = 1.928 \times 10^{24} \mathrm{ kg

Step 3: Gravitational field strength

g = \frac{GM_{\mathrm{enc}}{r^2} = \frac{6.67 \times 10^{-11} \times 1.928 \times 10^{24}}{(4.371 \times 10^6)^2}

= \frac{1.286 \times 10^{14}}{1.911 \times 10^{13}} = 6.73 \mathrm{ m/s^2

This is about $69\%$ of the surface gravity (9.81 \mathrm{ m/s^2), even though we are $31\%$ of the Way to the centre. Inside a uniform sphere, $g$ decreases linearly with $r$.

Board-Specific Content: SQA Advanced Higher

Gravitational Potential

The gravitational potential at distance $r$ from a point mass $M$ is:

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

This is the energy per unit mass. It is always negative (zero at infinity) and becomes more negative Closer to the mass.

Example: Calculate the gravitational potential at the surface of Earth and at an altitude of 300 \mathrm{ km.

Surface: V = -\frac{6.67 \times 10^{-11} \times 5.97 \times 10^{24}}{6.371 \times 10^6} = -6.25 \times 10^7 \mathrm{ J/kg

At 300 \mathrm{ km: V = -\frac{6.67 \times 10^{-11} \times 5.97 \times 10^{24}}{6.671 \times 10^6} = -5.97 \times 10^7 \mathrm{ J/kg

The potential difference: \Delta V = (-5.97 + 6.25) \times 10^7 = 2.8 \times 10^6 \mathrm{ J/kg.

Additional Practice Problems

26. Calculate the escape velocity from the surface of Mars. (Mars mass = 6.39 \times 10^{23} \mathrm{ kgMars radius = 3.39 \times 10^6 \mathrm{ m.) Compare it to the escape velocity from Earth.

27. A satellite in an elliptical orbit has a perigee altitude of 300 \mathrm{ km and an apogee altitude of 3000 \mathrm{ km. Calculate the semi-major axis, the orbital period, and the speeds at perigee and apogee.

28. A star has a surface temperature of 12000 \mathrm{ K and a radius of $2.5 R_\odot$. Calculate its luminosity relative to the Sun (T_\odot = 5778 \mathrm{ K) and its absolute magnitude.

29. Explain the significance of the cosmic microwave background radiation. At what temperature is it currently measured, and what does this tell us about the age of the universe?

30. Calculate the gravitational potential energy of a 100 \mathrm{ kg satellite in a circular orbit at an altitude of 500 \mathrm{ km. What is its total energy (kinetic + potential)?