IB MYP 4-5 Physics- Role of gravitational attraction in astronomy- Study Notes - New Syllabus
IB MYP 4-5 Physics-Role of gravitational attraction in astronomy- Study Notes
Key Concepts
- Role of gravitational attraction in astronomy
Role of Gravitational Attraction
Role of Gravitational Attraction
Holding Stars in Galaxies
Galaxies are massive systems that include billions of stars, planets, gas clouds, dust, and dark matter. Despite this huge scale, they remain bound together due to gravity.
- Every star in a galaxy pulls on every other star through gravitational attraction. Even though the pull from a single star is small, the combined effect of billions of stars adds up to a very strong gravitational field that holds the galaxy together.
- Gravity prevents stars from drifting away into empty space. Without it, the stars would scatter apart because of their own motion.
- Stars within a galaxy move in orbits around the galaxy’s center. This center is usually dominated by a supermassive black hole that exerts an extremely powerful gravitational force, helping anchor the galaxy’s structure.
- The galaxy’s spiral arms or elliptical shape are determined by the balance between the gravitational pull inward and the stars’ orbital motion outward.
- Key point: Gravity acts as the “glue” of the galaxy, ensuring that stars remain in a common structure instead of being scattered randomly.
Keeping Planets in Orbit Around Stars
Planets orbit stars (like Earth orbits the Sun) because of the star’s gravitational pull. Without gravity, planets would simply move in a straight line due to inertia.
Newton’s Law of Gravitation explains that the gravitational force between two objects depends on their masses and the distance between them:
\( F = G \dfrac{m_1 m_2}{r^2} \)
where \(m_1\) and \(m_2\) are the masses of the objects, \(r\) is the distance between their centers, and \(G\) is the gravitational constant.
For planets:
- The star (Sun) provides the gravitational pull.
- The planet’s velocity provides the tendency to move forward (inertia).
- The balance between gravity pulling inward and inertia pushing outward keeps planets in orbit.
- Planets closer to the star (like Mercury) orbit faster because the gravitational pull is stronger at shorter distances. Planets farther away (like Neptune) orbit more slowly, as the gravitational pull is weaker.
- This explains Kepler’s Third Law: planets farther from the Sun take longer to complete one orbit.
- Gravity also keeps moons in orbit around planets. For example, the Moon orbits Earth due to Earth’s gravitational pull.
- Key point: Gravity acts as the centripetal force that bends a planet’s straight-line motion into a stable curved orbit.
Example
Find the average orbital speed of Earth, given the mean Sun–Earth distance \(r = 1\,\text{AU} = 1.496\times10^{11}\,\text{m}\) and orbital period \(T = 365.25\) days.
▶️ Answer / Explanation
Step 1 — Convert period to seconds: \(T = 365.25 \times 24 \times 3600 \;=\; 3.156\times10^{7}\,\text{s}.\)
Step 2 — Use circular speed formula: Orbital (tangential) speed \(v = \dfrac{2\pi r}{T}.\)
Step 3 — Substitute numbers: \(v = \dfrac{2\pi \times 1.496\times10^{11}}{3.156\times10^{7}} \approx 2.9786\times10^{4}\,\text{m/s}.\)
Final answer: \(\boxed{v \approx 2.98\times10^{4}\ \text{m/s} \;(\approx 29.8\ \text{km/s})}\)
Example
Calculate the gravitational force pulling Earth toward the Sun using Newton’s law of gravitation. Use: \(G = 6.674\times10^{-11}\,\text{N m}^2\text{/kg}^2\), \(M_{\odot}=1.989\times10^{30}\,\text{kg}\), \(m_{\oplus}=5.972\times10^{24}\,\text{kg}\), and \(r=1.496\times10^{11}\,\text{m}\).
▶️ Answer / Explanation
Step 1 — Newton’s law: \(F = G\dfrac{M_{\odot}\,m_{\oplus}}{r^{2}}.\)
Step 2 — Substitute numbers: \(F = (6.674\times10^{-11})\dfrac{(1.989\times10^{30})(5.972\times10^{24})}{(1.496\times10^{11})^{2}}.\)
Step 3 — Compute (rounded): \(F \approx 3.54\times10^{22}\ \text{N}.\)
Interpretation: This enormous force supplies the centripetal force that keeps Earth in its nearly circular orbit.
Final answer: \(\boxed{F \approx 3.5\times10^{22}\ \text{N}}\)
Example
Estimate the Sun’s orbital speed about the centre of the Milky Way. Use an approximate radius from the galactic centre \(r \approx 8.0\ \text{kpc} \approx 2.47\times10^{20}\,\text{m}\), and assume the mass enclosed inside this radius is \(M \approx 1.0\times10^{11}\,M_{\odot}\) (where \(M_{\odot}=1.989\times10^{30}\,\text{kg}\)). Use \(v=\sqrt{\dfrac{GM}{r}}\).
▶️ Answer / Explanation
Step 1 — Compute mass in kg: \(M = 1.0\times10^{11}\times 1.989\times10^{30} \approx 1.989\times10^{41}\,\text{kg}.\)
Step 2 — Use orbital speed formula for circular motion under gravity: \(v = \sqrt{\dfrac{G M}{r}}.\)
Step 3 — Substitute numbers (with \(G=6.674\times10^{-11}\)): \(v \approx \sqrt{\dfrac{6.674\times10^{-11}\times 1.989\times10^{41}}{2.47\times10^{20}}}.\)
Step 4 — Compute (rounded): \(v \approx 2.32\times10^{5}\ \text{m/s} \;(\approx 232\ \text{km/s}).\)
Interpretation: This speed is typical for stars near the Sun’s orbit and is a result of the combined gravitational pull from the mass (visible + dark matter) inside that radius.
Final answer: \(\boxed{v \approx 2.3\times10^{5}\ \text{m/s} \;(\approx 230\ \text{km/s})}\)