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Orbits and satellites questions
An orbit is gravity taking on the centripetal job, and everything else follows from that one substitution. Out of it come Kepler's period-radius law, the energy books of a satellite, escape velocity, and the single orbit that hangs over a fixed point on the equator.
19 original questions · 53 marks · the orbits and satellites notes · Gravitational fields
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A satellite moves in a circular orbit around a planet. State what provides the centripetal force on the satellite and give its direction.
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The gravitational attraction of the planet on the satellite (1); directed towards the centre of the planet (the centre of the orbit) (1).State the relationship between the orbital period T and the orbital radius r for satellites orbiting the same planet.
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T2 ∝ r3: the square of the orbital period is proportional to the cube of the orbital radius (1).State two features of a geostationary orbit.
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Any two of the following, one mark each (2): period of 24 hours, equal to the Earth's rotation period; orbit is circular and lies in the equatorial plane; satellite moves west to east, in the same sense as the Earth's rotation, so it stays above a fixed point on the equator.A satellite is moved from a low circular orbit to a higher circular orbit around the same planet. State what happens to its orbital speed and to its orbital period.
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The orbital speed decreases, since v = √(GM/r) (1); the orbital period increases (1).Explain why the orbital speed of a satellite in a circular orbit does not depend on the satellite's mass.
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In GMm/r2 = mv2/r the satellite's mass m cancels, leaving v = √(GM/r) (1).Satellite B orbits a planet at four times the orbital radius of satellite A. Determine the ratio of B's orbital period to A's.
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T2 ∝ r3, so TB/TA = 43/2 (1)
TB/TA = 8 (1)A satellite orbits a planet of mass 6.0 × 1024 kg in a circular orbit of radius 7.0 × 106 m. Calculate the orbital speed of the satellite.
G = 6.67 × 10−11 N m2 kg−2Mark scheme
GMm/r2 = mv2/r, so v = √(GM/r) (1)
v = √(6.67 × 10−11 × 6.0 × 1024/(7.0 × 106)) (1)
v = 7561 m s−1 (1)A satellite travels at 7561 m s−1 in a circular orbit of radius 7.0 × 106 m. Calculate its orbital period.
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T = 2πr/v = (2π × 7.0 × 106)/7561 (1)
T = 5817 s (about 97 minutes) (1)Calculate the radius of a geostationary orbit around a planet of mass 6.0 × 1024 kg that rotates with a period of 24 hours (86 400 s).
Compare a low Earth orbit with a geostationary orbit. Refer to altitude, period and one typical use of each.
By treating the gravitational force as the centripetal force, derive the relationship T2 = 4π2r3/GM for a satellite in a circular orbit of radius r about a planet of mass M.
A moon of a distant planet orbits at a radius of 4.0 × 108 m with a period of 27 days. A second moon of the same planet orbits at a radius of 1.0 × 108 m. Determine the orbital period of the second moon, in days.
Show that the escape velocity from a point at distance r from a planet's centre is √2 times the speed of a satellite in a circular orbit of radius r.
A satellite of mass 500 kg orbits a planet of mass 6.0 × 1024 kg in a circular orbit of radius 8.0 × 106 m. Calculate the kinetic energy, the gravitational potential energy and the total energy of the satellite.
A moon orbits a planet of mass 6.0 × 1024 kg with an orbital period of 1.0 × 105 s. Determine the radius of its orbit.
Explain why a geostationary satellite must orbit directly above the equator.
An astronomer measures the orbits of four moons of a newly discovered planet:
moon W: r = 1.0 × 108 m, T = 2.0 days
moon X: r = 2.0 × 108 m, T = 5.7 days
moon Y: r = 4.0 × 108 m, T = 16.0 days
moon Z: r = 3.0 × 108 m, T = 9.0 days
The astronomer suspects that one measurement is wrong. Deduce which moon's data are inconsistent with the others.A satellite in a low orbit experiences a small drag force from the outer atmosphere. Explain why the satellite's speed increases as a result of the drag, even though its total energy decreases.
A satellite of mass 600 kg is to be moved from a circular orbit of radius 9.0 × 106 m to a circular orbit of radius 1.8 × 107 m around a planet of mass 6.0 × 1024 kg. Calculate the minimum energy that must be supplied.
G = 6.67 × 10−11 N m2 kg−2
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