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╝и╕╕╝TAP 403-1: Worked examples Ц Orbital Motion
Student sheet
Data required:
G = 6.67 ( 10-11 N m2 kg-2
mass of the Earth = 6.0 ( 1024 kg
radius of the Earth = 6.4 ( 106 m
mass of Jupiter = 1.9 ( 1027 kg
radius of Jupiter = 7.2 ( 107 m
JupiterТs day length = 10 hours
1) What is the only force acting on a single planet orbiting a star? Write down an expression for this force. If the planet moves in a circular orbit of radius r, at constant speed v, write down an expression for this speed in terms of the period T of the orbit.
Because the orbit is circular, the planet must experience a centripetal force of size mv2/r. Use this fact and the 2 expressions you have written down to prove KeplerТs third law, which states that the square of the time period of the planetТs orbit is proportional to the cube of the radius of the orbit.
2) Use KeplerТs third law, T2 ( r3, to answer this question. Two Earth satellites, A and B, orbit at radii of 7.0 ( 106 m and 2.8 ( 107 m respectively. Which satellite has the longer period of orbit? What is the ratio of orbital radii for the two satellites? What, therefore, is the ratio of the cubes of the orbital radii? What, therefore, is the ratio of the squares of the orbital periods? Finally therefore, what is the ratio of the satellitesТ orbital periods?
3) What is a geostationary satellite? Describe and explain the orbit of such a satellite. What might such a satellite be used for? With the help of your final expression in question 1, work out the orbital radius of such a satellite. What height is this above the EarthТs surface?
4) Suppose we wanted to place a satellite in Уjovi-stationaryФ orbit around Jupiter (the same as geostationary, but around Jupiter, not Earth). What orbital period would it need? What orbital radius would this correspond to?
Orbital Motion Ц TeacherТs Sheet
Data required:
G = 6.67 ( 10-11 N m2 kg-2
mass of the Earth = 6.0 ( 1024 kg
radius of the Earth = 6.4 ( 106 m
mass of Jupiter = 1.9 ( 1027 kg
radius of Jupiter = 7.2 ( 107 m
JupiterТs day length = 10 hours
1) What is the only force acting on a single planet orbiting a star? Write down an expression for this force. If the planet moves in a circular orbit or radius r, at constant speed v, write down an expression for this speed in terms of the period T of the orbit.
Because the orbit is circular, the planet must experience a centripetal force of size mv2/r. Use this fact and the 2 expressions you have written down to prove KeplerТs third law, which states that the square of the time period of the planet s orbit is proportional to the cube of the radius of the orbit.
Gravitational attraction GMm/r2.
v = 2pЁr/T
GMm/r2 = mv2/r
GM/r = m (2pЁr/T)2
GM/r = 4pЁ2r2m/T2
T2 = (4pЁ2/GM) r3
2) Use Kepler s third law, T2 ( r3, to answer this question. Two Earth satellites, A and B, orbit at radii of 7.0 ( 106 m and 2.8 ( 107 m respectively. Which satellite has the longer period of orbit? What is the ratio of orbital radii for the two satellites? What, therefore, is the ratio of the cubes of the orbital radii? What, therefore, is the ratio of the squares of the orbital periods? Finally therefore, what is the ratio of the satellitesТ orbital periods?
By KeplerТs third law, orbital period increases with orbital radius. Thus B has the longer orbital period.
rB/rA = 2.8 x 107/ 7.0 x 106 = 4
(rB/rA)3 = 43 = 64
By KeplerТs third law, (TB/TA)2 = (rB/rA)3 = 64
Thus TB/TA = square root of 64 = 8
3) What is a geostationary satellite? Describe and explain the orbit of such a satellite. What might such a satellite be used for? With the help of your final expression in question 1, work out the orbital radius of such a satellite. What height is this above the EarthТs surface?
A satellite that appears to be stationary over a point on the EarthТs equator. The orbit of such a satellite is circular and over the EarthТs equator as the satelliteТs orbital centre is the centre of the Earth, and the only points on Earth that orbit its centre are those on the equator.
Such a satellite might be used for communications, e.g. satellite broadcasting.
T2 = (4pЁ2/GM) r3
r3 = (GM/4pЁ2) T2
= (6.67 x 10-11 x 6.0 x 1024 / 4pЁ2) x (24 x 60 x 60)2
= 7.567 x 1022 m3 (4sf)
Therefore, r = 4.23 x 107 m (3sf)
Height above Earth s surface is 4.23 x 107 6.4 x 106 = 3.59 x 107 m or 36,000 km.
4) Suppose we wanted to place aKLTUX[_acf~ВЕвгжи└┴─╟туцчм н ┴ ┬
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00&0*0204060Z0\0`0f0z0~0В0Д0Т0┬0─0╪0·Ўю·Ў·Ў·Ўю·Ў·Ў·Ў·Ўю·Ў·Ў·Ў·Ў·Ў·Ў·Ў·ЎьЎ·Ўю·Ў·Ўф┘ь┘═┘═┘ю═┘═┘ф┘═┘═┘Ў·Ўh:Chе5H*mHsHh:Chе5mHsHhе5mHsHUhе5OJQJhе5 hе5H*B satellite in jovi-stationary orbit around Jupiter (the same as geostationary, but around Jupiter, not Earth). What orbital period would it need? What orbital radius would this correspond to?
10 hours
r3 = (GM/4pЁ2) T2 (see previous question)
= (6.67 x 10-11 x 1.9 x 1027 / 4pЁ2) x (10 x 60 x 60)2
= 4.160 x 1024 m3 (4sf)
Therefore, r = 1.61 x 108 m (3sf)
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