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S6 Physics: Circular Motion and Gravitation – Practice Paper 2 (Intermediate, 10 Questions)
S6 Physics: Circular Motion and Gravitation – Practice Paper 2 (Intermediate, 10 Questions)
Paper two concentrates on gravitation and orbits: the force between two masses, orbital speed from a period, GM from surface gravity, radius against period, the angular speed and radius of a geostationary orbit, why astronauts feel weightless, and the derivation of orbital speed from equating gravity to the required centripetal force. The last item takes apart the claim that higher orbits are faster.
Quick answer The force between the two masses is about 8.34 times 10 to the minus 10 N; a 5800 s period at 7.0 times 10 to the 6 m gives 7.6 km/s; GM is g R squared, about 4.0 times 10 to the 14; a larger radius gives a longer period; a geostationary satellite has angular speed about 7.3 times 10 to the minus 5 rad/s and a radius near 4.2 times 10 to the 7 m; weightlessness means no support force; four radii give eight times the period; orbital speed is the square root of GM over r, about 7.6 km/s; a higher orbit is slower.
01 Paper details (check answers afterwards)
| Item | Detail |
|---|---|
| Questions | 10 questions |
| Suggested time | about 28 minutes |
| Difficulty mix | Foundation 4 · Intermediate 3 · Challenge 2 · Error redo 1 |
| Chapter | Chapter: S6 Physics | Circular Motion and Gravitation: Centripetal Force, Angular Speed and Orbital Period |
| Level | S6 |
02 Questions (10 items)
- FoundationTwo masses of 5 kg and 10 kg are 2 m apart. Find the gravitational force between them.
- FoundationA satellite orbits at radius 7.0 times 10 to the 6 m with a period of 5.8 times 10 cubed s. Find its orbital speed.
- FoundationGiven surface gravity 9.8 m/s squared and Earth radius 6.4 times 10 to the 6 m, find GM for the Earth.
- FoundationFor satellites around the Earth, does a larger orbital radius give a longer or a shorter period?
- IntermediateA geostationary satellite has a period of 24 hours. Find its angular speed.
- IntermediateUsing GM about 4.0 times 10 to the 14, estimate the orbital radius of a geostationary satellite.
- IntermediateWhy do astronauts on an orbiting space station feel weightless?
- ChallengeA planet orbits at four times the Earth orbital radius. How many times the Earth period is its period?
- ChallengeDerive the circular orbit speed from GMm over r squared equals mv squared over r, then evaluate it at r = 7.0 times 10 to the 6 m.
- Error redoA student says a higher orbit means a faster satellite, so higher orbits are easier to reach. What is wrong?
03 Answers and one-line notes
| # | Answer | One-line note |
|---|---|---|
| 1 | F = G m1 m2 over r squared = 6.67 times 10 to the minus 11 times 50 divided by 4, about 8.34 times 10 to the minus 10 N | Gravitational force grows with the product of the masses and falls as the square of the distance. |
| 2 | v = 2 pi r over T = 2 pi times 7.0 times 10 to the 6 divided by 5800, about 7.6 times 10 cubed m/s, that is 7.6 km/s | Orbital speed is again the circumference divided by the period. |
| 3 | GM = g R squared = 9.8 times 6.4 times 10 to the 6 squared, about 4.0 times 10 to the 14 m cubed per s squared | GM is a fixed constant for the Earth and is used in every orbit calculation. |
| 4 | A longer period; Kepler third law gives T squared proportional to r cubed, so a larger radius means a longer period | A higher orbit is slower, and both facts come from the same law. |
| 5 | T = 86400 s; angular speed = 2 pi over 86400, about 7.3 times 10 to the minus 5 rad/s | A geostationary satellite must have the same angular speed as the Earth rotation. |
| 6 | r cubed = GM T squared over 4 pi squared, which is about 7.6 times 10 to the 22, so r is about 4.2 times 10 to the 7 m | Work out r cubed first and then take the cube root, which gives about 42 000 km. |
| 7 | The astronaut and the station fall freely with the same acceleration, so there is no normal force between them and no sensation of support; gravity has not disappeared | Weightlessness means no support force, not the absence of gravity. |
| 8 | Divide the cubes: the period ratio squared is 4 cubed, which is 64, so the period ratio is 8 | Square the ratio first and then take the square root instead of answering 4. |
| 9 | Cancelling gives v = square root of GM over r, which is about 7.6 times 10 cubed m/s | The mass m cancels, so the orbital speed does not depend on the satellite mass. |
| 10 | Since v is the square root of GM over r, a larger r gives a smaller speed; a higher orbit is slower, though it costs more energy to lift the satellite there | Higher means faster only for the period, not for the speed. |
04 What to do after marking
- After marking, redo only the wrong items and write down which step went wrong (formula, unit, or misread number).
- Repeat the same item two days later; move to the next paper only when both attempts are correct.
- Parents only need the answer and note columns; there is no need to explain every item.
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