krit.club logo

Physics - Space Physics

Grade 10IGCSE

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

🔑Concepts

•

The Earth rotates on its axis once every 2424 hours, resulting in the day-night cycle, and orbits the Sun in approximately 365.25365.25 days.

•

The Moon orbits the Earth every 27.327.3 days; however, the lunar month (from new moon to new moon) is approximately 29.529.5 days.

•

The Sun is a medium-sized star that releases energy through nuclear fusion, where hydrogen nuclei fuse to form helium: 4 11H→ 24He+2 +10e+energy4\,^{1}_{1}H \rightarrow \,^{4}_{2}He + 2\,^{0}_{+1}e + \text{energy}.

•

Orbital speed vv is determined by the distance from the center of the object being orbited rr and the orbital period TT.

•

The Solar System contains eight planets: Mercury, Venus, Earth, Mars, Jupiter, Saturn, Uranus, and Neptune (MVEMJSUN). All orbit the Sun in elliptical paths.

•

Stars are formed from interstellar clouds of dust and gas (nebulae) collapsing under gravity to form a protostar.

•

A star's lifecycle depends on its mass: Stable stars undergo fusion in the 'Main Sequence' stage before expanding into a Red Giant (low mass) or Red Supergiant (high mass).

•

Redshift is the observed increase in the wavelength λ\lambda of light from distant galaxies, indicating they are moving away from Earth.

•

Hubble's Law states that the recession velocity vv of a galaxy is directly proportional to its distance dd from Earth: v=H0dv = H_0 d.

•

The Cosmic Microwave Background Radiation (CMBR) provides evidence for the Big Bang Theory, representing the 'afterglow' of the early hot universe.

📐Formulae

v=2πrTv = \frac{2\pi r}{T}

v=H0dv = H_0 d

d=vH0d = \frac{v}{H_0}

t≈1H0t \approx \frac{1}{H_0}

💡Examples

Problem 1:

A satellite orbits the Earth at a constant altitude. The radius of the orbit is 7.0×106 m7.0 \times 10^6\text{ m} and the time taken for one complete orbit is 90 minutes90\text{ minutes}. Calculate the orbital speed vv in m/s\text{m/s}.

Solution:

T=90×60=5400 sT = 90 \times 60 = 5400\text{ s} v=2π×(7.0×106)5400≈8145 m/sv = \frac{2 \pi \times (7.0 \times 10^6)}{5400} \approx 8145\text{ m/s}

Explanation:

First, convert the orbital period from minutes to seconds. Then apply the orbital speed formula v=2πrTv = \frac{2\pi r}{T} using the orbital radius in meters.

Problem 2:

A distant galaxy is observed to be moving away from Earth at a speed of 3.3×106 m/s3.3 \times 10^6\text{ m/s}. Given that Hubble's constant H0H_0 is 2.2×10−18 s−12.2 \times 10^{-18}\text{ s}^{-1}, calculate the distance dd to the galaxy in meters.

Solution:

d=vH0d = \frac{v}{H_0} d=3.3×1062.2×10−18=1.5×1024 md = \frac{3.3 \times 10^6}{2.2 \times 10^{-18}} = 1.5 \times 10^{24}\text{ m}

Explanation:

Using Hubble's Law v=H0dv = H_0 d, rearrange to solve for dd. This shows the vast distances involved in extra-galactic astronomy.

Problem 3:

Estimate the age of the Universe in years given Hubble's constant H0=2.2×10−18 s−1H_0 = 2.2 \times 10^{-18}\text{ s}^{-1}.

Solution:

t=1H0=12.2×10−18≈4.54×1017 st = \frac{1}{H_0} = \frac{1}{2.2 \times 10^{-18}} \approx 4.54 \times 10^{17}\text{ s} Age in years=4.54×1017365.25×24×3600≈1.44×1010 years\text{Age in years} = \frac{4.54 \times 10^{17}}{365.25 \times 24 \times 3600} \approx 1.44 \times 10^{10}\text{ years}

Explanation:

The age of the universe is approximately the reciprocal of the Hubble constant. To find the value in years, divide the result in seconds by the number of seconds in a year (3.15×107 s3.15 \times 10^7\text{ s}).

Space Physics - Revision Notes & Key Formulas | IGCSE Grade 10 Science