Cambridge IGCSE Physics · topic 6.1
The Earth and the Solar System
Connect the ideas, make your working visible, and test the result. Use the hints when you need a starting point; open the full solution after your attempt.
Extra challenging questions
Work through the linked stages and explain what each result means.
A satellite raised to twice the orbital radius
Optional enrichment: this challenge may extend beyond the usual syllabus depth and assumes confidence with algebra and graph interpretation. A spherical planet has radius R = 6.0×10⁶ m and surface gravitational field g₀ = 10 N kg⁻¹. A 100 kg satellite follows a circular orbit at radius 2R measured from the planet's centre. Ignore rotation and atmosphere. Use GM = g₀R² and the supplied inverse-square field data. Supplied model: Use g = GM/r², circular acceleration v²/r and orbital period T = 2πr/v. Supplied data (Radius / R; Field magnitude / N kg⁻¹): (1, 10), (2, 2.5), (3, 1.111111), (4, 0.625). (a) Find the satellite's weight, orbital speed and orbital period. Explain what supplies the centripetal resultant. [3 marks] (b) The radius is increased to 4R. Predict ratios of field, speed and period, and explain why astronauts can feel weightless despite nonzero gravity. [3 marks]
Planetary gravitational field
Supplied question data. Read the labelled axes and use the observations to test the physical model.
Vertical axis (y): Field magnitude / N kg⁻¹
Horizontal axis (x): Radius / R
Read the exact data as a table
| Radius / R | Field magnitude / N kg⁻¹ |
|---|---|
| 1 | 10 |
| 2 | 2.5 |
| 3 | 1.111111 |
| 4 | 0.625 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Begin with part (a): Find the satellite's weight, orbital speed and orbital period. Explain what supplies the centripetal resultant.
Reveal the worked solution
- At 2R the field is g₀/4 = 2.5 N kg⁻¹, so the gravitational force is mg = 250 N. The satellite's mass remains 100 kg.
- Gravity alone supplies mv²/r: v = √(gr) = √(2.5×1.2×10⁷)≈5477 m s⁻¹.
- Period is circumference/speed, T = 2πr/v≈1.38×10⁴ s. 'Centripetal force' names the inward resultant; it is not an additional interaction alongside gravity.
- For a circular orbit g∝r⁻², v∝r⁻¹/² and T∝r³/². Doubling r therefore gives the stated ratios.
- Astronauts and spacecraft accelerate together in free fall. Little support force acts on the astronauts, so their apparent weight can be near zero.
- The gravitational field is not zero in orbit, and constant speed still requires acceleration because velocity direction changes.
Answer (a) Weight = 250 N; orbital speed ≈5.48 km s⁻¹; period ≈1.38×10⁴ s. (b) Field becomes 1/4, speed 1/√2 and period 2√2 times the initial orbital values.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
A scale model that cannot show every scale
Optional enrichment: this challenge may extend beyond the usual syllabus depth and assumes confidence with algebra and graph interpretation. Use 1 AU = 1.50×10¹¹ m, solar radius = 7.0×10⁸ m and 1 parsec = 3.09×10¹⁶ m. A model puts Earth 15 m from the Sun. The supplied graph lists stellar parallax against distance; parallax p is the half-angle after Earth's six-month baseline shift, and d/pc = 1/(p/arcsec). A star has parallax 0.050 arcsec. Supplied data (Distance / pc; Parallax / arcsec): (5, 0.2), (10, 0.1), (20, 0.05), (40, 0.025). (a) Find the Sun's model radius and the star's true and model distances. Explain the practical scale limitation. [3 marks] (b) A parallax measurement has uncertainty ±0.005 arcsec. Estimate the fractional distance uncertainty and compare with a standard-candle method. [3 marks]
Parallax calibration
Supplied question data. Read the labelled axes and use the observations to test the physical model.
Vertical axis (y): Parallax / arcsec
Horizontal axis (x): Distance / pc
Read the exact data as a table
| Distance / pc | Parallax / arcsec |
|---|---|
| 5 | 0.2 |
| 10 | 0.1 |
| 20 | 0.05 |
| 40 | 0.025 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Begin with part (a): Find the Sun's model radius and the star's true and model distances. Explain the practical scale limitation.
Reveal the worked solution
- The model scale is 15/(1.50×10¹¹) = 1.0×10⁻¹⁰, so solar radius becomes 0.070 m.
- The parallax gives d = 1/0.050 = 20 pc, equal to 6.18×10¹⁷ m. Applying the same scale gives 6.18×10⁷ m, or 61800 km.
- A classroom model that fits planetary orbits cannot also put the nearest stellar distances nearby at the same scale; compressed diagrams must state their distortion.
- For d = 1/p, small fractional uncertainties satisfy Δd/d≈Δp/p = 0.005/0.050 = 0.10.
- The exact bounds are 1/0.055≈18.2 pc and 1/0.045≈22.2 pc, showing asymmetric inversion of the parallax limits.
- A standard candle uses known luminosity and measured flux with an inverse-square model. It reaches farther than practical parallax but requires a calibrated intrinsic luminosity and corrections for absorption.
Answer (a) Model Sun radius = 0.070 m; star distance = 20 pc = 6.18×10¹⁷ m; model star distance = 6.18×10⁷ m. (b) Approximate distance uncertainty = 10%, or about ±2 pc; exact bounds are about 18.2–22.2 pc.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.