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Questions on orbits and satellites
Gravity supplies the centripetal force in an orbit, and equating GMm/r² to mv²/r gives the orbital speed and Kepler's third law. A satellite's kinetic, potential and total energy follow from the same substitution, and escape velocity √(2GM/r) does not depend on the escaping mass. The geostationary orbit is circular and equatorial.
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 = 7.56 × 103 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 = 5.82 × 103 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).
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T2 = 4π2r3/GM, so r = (GMT2/4π2)1/3 (1)
r = ((6.67 × 10−11 × 6.0 × 1024 × 864002)/4π2)1/3 (1)
r = 4.23 × 107 m (1)Compare a low Earth orbit with a geostationary orbit. Refer to altitude, period and one typical use of each.
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Low orbit: altitude of a few hundred kilometres with a period of about 90 minutes; geostationary: altitude of about 36 000 km with a period of 24 hours (1). Low orbits are used for Earth imaging (or the space station), as they pass close to the surface (1). Geostationary orbits are used for communications or weather satellites, which must remain above a fixed point on the equator (1).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.
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GMm/r2 = mv2/r (gravitational force provides the centripetal force) (1)
Substitute v = 2πr/T: GM/r = 4π2r2/T2 (1)
Rearranging gives T2 = 4π2r3/GM (1)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.
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T2/r3 is the same for both moons (both orbit the same central mass) (1)
T2 = 27 × (1.0/4.0)3/2 (1)
T2 = 3.4 days (1)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.
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Orbital speed: vorb = √(GM/r), from gravity acting as the centripetal force (1)
Escape velocity: vesc = √(2GM/r) (1)
vesc/vorb = √2, whatever the value of r (1)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.
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v = √(GM/r) = 7.07 × 103 m s−1 (1)
KE = ½mv2 = 1.25 × 1010 J (1)
PE = −GMm/r = −2.50 × 1010 J (1)
Total = KE + PE = −1.25 × 1010 J (1)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.
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r = (GMT2/4π2)1/3 (1)
r = ((6.67 × 10−11 × 6.0 × 1024 × (1.0 × 105)2)/4π2)1/3 (1)
r = 4.66 × 107 m (1)Explain why a geostationary satellite must orbit directly above the equator.
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The plane of any orbit must contain the centre of the Earth (1). To remain above a fixed point as the Earth rotates, the satellite's orbital plane must coincide with the equatorial plane (1); an orbit inclined to the equator would carry the satellite north and south of the point, so it would not appear stationary (1).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.Mark scheme
For moons of one planet, T2/r3 should be the same for all (1)
In units of days2 per (108 m)3: W gives 4.0, X gives 4.1, Y gives 4.0 (1)
Z gives 3.0 (1)
W, X and Y agree; moon Z's data are inconsistent with Kepler's third law, so Z's measurement is the suspect one (1).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.
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Drag transfers energy away from the satellite, so its total energy falls (1); the satellite therefore spirals down to a smaller orbital radius (1); the orbital speed v = √(GM/r) increases as r decreases, so the kinetic energy rises (1); the potential energy falls by more than the kinetic energy rises, giving the net loss of total energy (1).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−2Mark scheme
In a circular orbit KE = ½mv2 = GMm/2r and PE = −GMm/r, so the total energy is E = −GMm/2r (1)
E1 = −GMm/(2 × 9.0 × 106) = −1.33 × 1010 J (1)
E2 = −GMm/(2 × 1.8 × 107) = −6.67 × 109 J (1)
Energy supplied = E2 − E1 (1)
= 6.67 × 109 J (1)
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