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Space Physics - Stars and the Universe

Grade 12IGCSEPhysics

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

Stars are formed from interstellar clouds of dust and gas called nebulae. Gravity pulls the matter together to form a protostar, which increases in temperature until nuclear fusion of hydrogen into helium begins (411H24He+2+10e+energy4 \, ^{1}_{1}\text{H} \rightarrow \, ^{4}_{2}\text{He} + 2 \, ^{0}_{+1}e + \text{energy}), marking the start of the Main Sequence phase.

The stable period of a star (Main Sequence) is a result of the balance between the inward force of gravity and the outward pressure from nuclear fusion (hydrostatic equilibrium).

Stars with a mass similar to the Sun eventually evolve into Red Giants, then shed their outer layers to leave behind a hot, dense core called a White Dwarf.

Stars with much higher mass than the Sun evolve into Red Supergiants, which eventually explode in a Supernova, leaving behind either a Neutron Star or a Black Hole.

The orbital speed vv of a planet or satellite in a circular orbit of radius rr is calculated using the period TT: v=2πrTv = \frac{2\pi r}{T}.

Redshift is the observed increase in the wavelength of light from distant galaxies. The change in wavelength Δλ\Delta \lambda relative to the laboratory wavelength λ0\lambda_0 is proportional to the recessional velocity vv of the galaxy: Δλλ0vc\frac{\Delta \lambda}{\lambda_0} \approx \frac{v}{c}.

Hubble’s Law states that the recessional velocity vv of a galaxy is proportional to its distance dd from Earth, expressed as v=H0dv = H_0 d, where H0H_0 is the Hubble constant.

The age of the universe can be estimated using the reciprocal of the Hubble constant: t1H0t \approx \frac{1}{H_0}. Current estimates put this at approximately 13.813.8 billion years (4.35×1017 s4.35 \times 10^{17} \text{ s}).

📐Formulae

v=2πrTv = \frac{2\pi r}{T}

v=H0dv = H_0 d

z=Δλλ0=λobsλemitλemitz = \frac{\Delta \lambda}{\lambda_0} = \frac{\lambda_{obs} - \lambda_{emit}}{\lambda_{emit}}

Δλλ0=vc\frac{\Delta \lambda}{\lambda_0} = \frac{v}{c}

t1H0t \approx \frac{1}{H_0}

💡Examples

Problem 1:

A galaxy is moving away from Earth at a speed of 3.5×104 km/s3.5 \times 10^4 \text{ km/s}. Given the Hubble constant H0=2.2×1018 s1H_0 = 2.2 \times 10^{-18} \text{ s}^{-1}, calculate the distance to the galaxy in meters.

Solution:

d=vH0=3.5×107 m/s2.2×1018 s1=1.59×1025 md = \frac{v}{H_0} = \frac{3.5 \times 10^7 \text{ m/s}}{2.2 \times 10^{-18} \text{ s}^{-1}} = 1.59 \times 10^{25} \text{ m}

Explanation:

We use Hubble's Law v=H0dv = H_0 d. First, convert the velocity from km/s\text{km/s} to m/s\text{m/s} by multiplying by 10310^3. Then, divide the velocity by the Hubble constant to find the distance.

Problem 2:

Light from a distant galaxy has a measured wavelength of 670 nm670 \text{ nm} for a specific hydrogen line. The laboratory wavelength for this line is 656 nm656 \text{ nm}. Calculate the speed at which the galaxy is moving away from Earth. (Speed of light c=3.0×108 m/sc = 3.0 \times 10^8 \text{ m/s})

Solution:

Δλ=670 nm656 nm=14 nm\Delta \lambda = 670 \text{ nm} - 656 \text{ nm} = 14 \text{ nm} vc=Δλλ0    v=c×14 nm656 nm\frac{v}{c} = \frac{\Delta \lambda}{\lambda_0} \implies v = c \times \frac{14 \text{ nm}}{656 \text{ nm}} v=3.0×108×0.0213=6.4×106 m/sv = 3.0 \times 10^8 \times 0.0213 = 6.4 \times 10^6 \text{ m/s}

Explanation:

The change in wavelength Δλ\Delta \lambda is found first. Using the redshift formula z=Δλλ0z = \frac{\Delta \lambda}{\lambda_0}, we find the ratio of the change. Multiplying this ratio by the speed of light cc gives the recessional velocity vv.

Stars and the Universe - Revision Notes & Key Formulas | IGCSE Grade 12 Physics