IGCSE Physics 0625 · Section 6 · Space Physics — Numerical Solutions
Space Physics — Numerical Practice: Solutions
A. Orbital Speed
A1
r=7×10⁶m, T=5800s. Find orbital speed.
1
v = 2πr/T
(2 × π × 7×10⁶) ÷ 5800
✓≈ 7583 m/s
A2
r=4×10⁵m, T=7.2×10⁴s. Find speed.
1
v = 2πr/T
(2 × π × 4×10⁵) ÷ 7.2×10⁴
✓≈ 34.9 m/s
A3
T=3.15×10⁷s, r=1.5×10¹¹m. Find speed.
1
v = 2πr/T
(2 × π × 1.5×10¹¹) ÷ 3.15×10⁷
✓≈ 29900 m/s (≈ 29.9 km/s — this matches Earth’s real orbital speed around the Sun)
A4
v=3000 m/s, r=8×10⁶m. Find period.
1
Rearrange v=2πr/T to T=2πr/v
(2 × π × 8×10⁶) ÷ 3000
✓≈ 16755 s (≈ 4.65 hours)
A5
Why is a further planet’s orbital speed slower?
✓The Sun’s gravitational field strength decreases with distance. A weaker gravitational pull provides less centripetal force to hold a planet in a fast orbit, so more distant planets settle into slower orbital speeds.
B. Light Travel Time and Light-Years
B1
Earth-Sun distance 1.5×10¹¹m. Find light travel time.
1
time = distance/speed
(1.5×10¹¹) ÷ (3×10⁸)
✓500 s (≈ 8.3 minutes)
B2
Star 4×10¹⁶m away. Find time in s, then years.
1
time = distance/speed
(4×10¹⁶) ÷ (3×10⁸) = 1.333×10⁸ s
2
Convert to years
(1.333×10⁸) ÷ (3.15×10⁷)
✓1.333×10⁸ s ≈ 4.23 years
B3
6.5 light-years to metres.
1
distance = number of ly × 9.5×10¹⁵
6.5 × 9.5×10¹⁵
✓6.175 × 10¹⁶ m
B4
Convert 2×10¹⁷m to light-years.
1
light-years = distance ÷ 9.5×10¹⁵
(2×10¹⁷) ÷ (9.5×10¹⁵)
✓≈ 21.1 light-years
B5
Moon 3.84×10⁸m away. Find light travel time.
1
time = distance/speed
(3.84×10⁸) ÷ (3×10⁸)
✓1.28 s
B6
Why use light-years, not metres?
✓Distances to stars are so vast that expressing them in metres would require extremely large, unwieldy numbers. The light-year gives a more manageable, intuitive scale for astronomical distances.
C. The Hubble Constant and the Expanding Universe
C1
v=5×10⁶m/s, H₀=2.2×10⁻¹⁸/s. Find distance.
1
d = v/H₀
(5×10⁶) ÷ (2.2×10⁻¹⁸)
✓≈ 2.27 × 10²⁴ m
C2
d=3×10²⁴m, H₀=2.2×10⁻¹⁸/s. Find speed.
1
v = H₀ × d
(2.2×10⁻¹⁸) × (3×10²⁴)
✓6.6 × 10⁶ m/s
C3
Estimate age of universe using H₀=2.2×10⁻¹⁸/s.
1
age ≈ 1/H₀
1 ÷ (2.2×10⁻¹⁸)
2
Convert to years
(4.545×10¹⁷) ÷ (3.15×10⁷)
✓≈ 4.55 × 10¹⁷ s ≈ 1.44 × 10¹⁰ years (≈ 14.4 billion years)
C4
What is redshift, and how does it support an expanding universe?
✓Redshift is an increase in the observed wavelength of light from a receding star or galaxy. Almost all distant galaxies show redshift when viewed from Earth, meaning they are moving away from us in every direction — exactly what would be expected if space itself is expanding, supporting the Big Bang theory.
C5
Why does d/v = 1/H₀ being constant support the Big Bang?
✓Since d/v gives the same value (1/H₀) regardless of which galaxy’s data is used, tracing every galaxy’s motion backward in time suggests they were all at the same place at the same moment in the past — consistent with all the matter in the Universe having been present at a single point, as the Big Bang theory describes.
Mastery Test — Answer Key
- ≈ 6981 m/s — v = 2π×5×10⁶ ÷ 4500
- 300 s (9×10¹⁰ ÷ 3×10⁸)
- 2.85 × 10¹⁶ m (3 × 9.5×10¹⁵)
- 2 light-years (1.9×10¹⁶ ÷ 9.5×10¹⁵)
- 4 × 10²⁴ m — d = v/H₀ = 8.8×10⁶ ÷ 2.2×10⁻¹⁸
- 1.1 × 10⁷ m/s — v = H₀×d = 2.2×10⁻¹⁸ × 5×10²⁴
- The Sun’s (or a galaxy’s local) gravitational field strength decreases with distance. A weaker gravitational pull provides less force to hold a planet in a fast, tight orbit, so orbital speed decreases further out — the same underlying idea (weaker field, less force, less speed) recurs throughout orbital mechanics.
- CMBR is microwave radiation observed from every direction in space, at a very specific characteristic frequency. It is believed to be the “afterglow” of radiation produced shortly after the Universe began, since stretched (redshifted) into the microwave region by billions of years of cosmic expansion — its existence and uniformity strongly support the Big Bang theory.
- Approximately 100,000 light-years.
- Light travels at a very fast but finite speed (3×10⁸ m/s). Since the Sun is about 1.5×10¹¹ m from Earth, even at this speed it takes a measurable time (distance ÷ speed ≈ 500 s ≈ 8.3 minutes) to cross that distance.
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