×
( . 22 10 m s× 14 3 –2
The two values agree within ~1%.
Extension: (Hard) Can you think of a reason for the ~1% discrepancy? (Hint: See Physics Phile ‘Finding new planets’ page 40.)
Kepler’s laws and satellites
Johannes Kepler (1571–1630), using Tycho Brahe’s data, showed that the known planets and Earth orbit the Sun in ellipses that obey his three laws of planetary motion, but he had no idea why. An ellipse is a circle, stretched along one dimension. The more stretched the ellipse, the more eccentric it is. Most of the orbits of planets of our solar system are very nearly circular, not very eccentric.
Comets have highly eccentric orbits.
Figure 2.2.3 defines some properties of elliptical orbits. The semimajor axis is half the length of the ellipse’s longest axis. It can also be thought of as a kind of average radius for the orbit. For a circular orbit, the semimajor axis is the radius.
The point of closest approach of the orbit to the central body is called the periapsis. The furthest point is the apoapsis. If the central body is the Sun, then these points can also be called perihelion and aphelion (helios is ‘Sun’ in Greek).
If the central body is the Earth, then they are the perigee and apogee (geos is
‘Earth’ in Greek).
Let a be the semimajor axis and let dA and dP be the distances from the central body to the aphelion and perihelion respectively, then look at Figure 2.2.3a and confirm that a is the average of these: a = (dA + dP)/2.
You had a sneak preview of Kepler’s laws of planetary motion in the Preliminary text (see in2 Physics @ Preliminary p 250). Here they are again:
1 The orbits of the planets are ellipses, with the Sun at one focus (Figure 2.2.3a).
Solve problems and analyse information using:
F Gm m
= d1 22
Solve problems and analyse information to calculate the centripetal force acting on a satellite undergoing uniform circular motion about the Earth using:
F mv
= r 2
Analyse the forces involved in uniform circular motion for a range of objects, including satellites orbiting the Earth.
means the planet travels faster when it is closer to the Sun (Figure 2.2.3b).
3 For all planets orbiting the Sun, the square of the orbital period T is proportional to the cube of the semimajor axis a. This is the law of periods.
T a
2
3 = constant constant
These laws hold for any orbiting system of two bodies if the mass of the central body is very much larger than the mass of the other. They also apply to circular orbits, since a circle is a special case of an ellipse.
If c is the half the distance between the foci of an ellipse, then eccentricity e is defined as e c
= . A circle is an ellipse with zero eccentricity (e = 0); both foci are a together at the circle centre. For a circle, the semimajor axis becomes the radius, a = r. Orbits of planets in our solar system are very nearly circular, the two most eccentric being Mercury with e = 0.2056 and Mars with e = 0.0934.
Pluto has e = 0.2482 but, sadly for Pluto-fans, it was demoted to a dwarf planet in 2006.
Sun
focus focus focus
perihelion (closest to the Sun)
aphelion (farthest from Sun)
Area = Area semimajor axis
a b
Figure 2.2.3 (a) A highly eccentric elliptical orbit. (b) Kepler’s law of areas: a line joining the planet to the Sun sweeps out equal areas in equal times.
Discuss the importance of Newton’s Law of Universal Gravitation in understanding and calculating the motion of satellites.
Newton derived Kepler’s laws from his laws of motion and gravitation. He showed that Kepler’s first law (law of elliptical orbits) follows from the inverse square law. By including his three laws of motion, he also proved Kepler’s second law (law of equal areas). These derivations are beyond the syllabus; however, showing Kepler’s third law, the law of periods, for a circular orbit is very easy.
Careful! Don’t confuse the semimajor axis a with centripetal acceleration ac. Suppose a satellite of mass m orbits a central body of mass M and m << M so that the acceleration of M is negligible. In a circular orbit, the satellite’s acceleration is centripetal:
a v
r v r
T a r
c= 2 and = 2π ∴ c= 4Tπ22
where v is orbital velocity, r is orbital radius and T is orbital period.
The magnitude of gravitational force exerted on m is:
F = mac = GmM ⇒ a GM
r
r T
T
r GM
c= 2 = 4π22 ∴ 32 =4π (which is constant)2 r 2
We derived this for a circular orbit, but it is also true for elliptical orbits if we replace radius r with semimajor axis a.
T
Because here the value of Kepler’s constant is explicit, we will call this the
‘explicit’ form of Kepler’s law of periods.
Worked example QUESTIon
From Earth, you observe (almost edge on) the orbit of Jupiter’s moon Ganymede and determine its orbital period to be T = 7.15 Earth days. You measure the width of the orbit to be w = 2.14 × 109 m. Assuming the orbit is circular, determine Jupiter’s mass.
SoLUTIon
Assuming orbit is circular: r w
= =2 1 07 10. × 9m
Explicit form of Kepler’s third Law: r T
An early triumph of the law of universal gravitation occurred when Newton’s friend Edmund Halley used it to show that the trajectory (Figure 2.2.4) of a comet he had observed (now called Halley’s Comet) fitted with the trajectories of two previously observed comets. Halley concluded it was the same comet, and correctly predicted that it would return every 76 years.
The success of Newton’s law of universal gravitation was not simply that it could be used to explain Kepler’s laws, which were already known, but that it could be used to predict other phenomena not yet observed (such as space travel). It could also be used to accurately predict small deviations of planets from Kepler’s ideal orbits around the Sun. For example, when planets pass near each other, local effects of gravity perturb them from perfect Keplerian orbits. The law of universal gravitation can be used to predict these deviations very accurately.
Halley included the effects of perturbations due to planets in his comet calculations. Similar deviations in the orbit of Uranus were attributed to the gravity of a then unknown planet. Neptune, that new planet, was found in 1846 within 1° of the position predicted using the law of universal gravitation. Several astronomers contributed to both the calculations and the observations, resulting in arguments about who deserved credit for discovering Neptune.
The first obvious failure of Newton’s gravitation law was in explaining the observation that the position of the perihelion of Mercury’s orbit was not fixed, but was precessing around the Sun. (See in2 Physics @ Preliminary p 255.) Perturbations due to gravity of other planets and other mechanical effects such as the Sun’s equatorial bulge were able to explain 99.23% of the precession, but the remaining 0.77% required an improved theory of gravitation—Einstein’s theory of general relativity.
Solve problems and analyse information using:
Figure 2.2.4 The eccentric orbit of Halley’s Comet in relation to the nearly circular orbits of Earth and other planets. The angle between the comet’s orbit and the plane of the orbits of the planets is not apparent here.
Worked example QUESTIon
Derive an expression for the magnitude of orbital velocity for a satellite in a circular orbit, in terms of mass of the central body M and orbital radius r. Use this expression to calculate the Moon’s orbital speed, assuming a circular orbit.
Data: Average Earth–Moon distance d = 3.84 × 108 m Earth’s mass Me = 5.97 × 1024 kg
SoLUTIon
Gravitational acceleration: a GM
g= r2 Resulting centripetal acceleration: a a v
r GM
g r
c= = 2= 2
Rearrange:
v GM
= r for circular orbits
Moon’s orbital speed:
v= × × ×
× =
− −
6 67 10 5 97 10
3 84 1011 8 24 1020 1
. .
. m s
Note that orbital speed is independent of the mass of the satellite. One of the consequences of this equation is that for circular orbits, the smaller the radius the faster the orbital speed.
Define the term orbital velocity and the quantitative and qualitative relationship between orbital velocity, the gravitational constant, mass of the central body, mass of the satellite and the radius of the orbit using Kepler’s Law of Periods.