__Kepler's third law__

Use Kepler's third law to calculate the mass of the sun, assuming that the
orbit of the earth around the sun is circular, with radius r = 1.5*10^{8}
km.

Solution:

- Concepts:

Kepler's third law - Reasoning:

mv^{2}/r = GMm/r^{2}. v^{2}= GM/r. (2πr/T)^{2}= GM/r.

T^{2}= (4π^{2}/(GM))r^{3}, Kepler's third law. - Details of the calculation:

T = 365 days = 3.15*10^{7}s.

M = (4π^{2}/(GT^{2}))r^{3}, M = 2.01*10^{30}kg.

Haley's Comet approaches the sun to within 0.570 A.U., and its orbital period
is 75.6 years. (A.U. is the abbreviation for astronomical units, where 1 A.U. =
1.5*10^{11}m is the mean Earth-Sun distance.) How far from the sun will
Haley's comet travel before it starts its return journey?

Solution:

- Concepts:

Motion in a central potential, Kepler's third law - Reasoning:

We are asked to find R_{max}, given R_{min}and the period T for a comet orbiting the sun. Since the mass of the sun (m_{1}) is much greater than the mass of the comet, we may consider the sun to be stationary. - Details of the calculation:

T^{2 }= (4π^{2}/Gm_{1})R^{3}.

For elliptical orbits, R denotes the semi-major axis, R = (R_{max }+ R_{min})/2.

R^{3 }= Gm_{1}T^{2}/4π^{2}.

R^{3 }= (6.67*10^{-11 }Nm^{2}/kg^{2})(1.991*10^{30 }kg)(75.6*365*24*60*60 s)^{2}/4π^{2 }= 1.91*10^{37 }m^{3}.

R = 2.67*10^{12 }m = (R_{max }+ R_{min})/2.

R_{max }= 5.35*10^{12 }m - 0.570*1.5*10^{11 }m = 5.26*10^{12 }m.

R_{max}is the maximum distance of the comet from the sun.

The time of revolution of planet Jupiter around the Sun is
T_{J} ~ 12 years. What is the distance between Jupiter and the Sun if
the Earth-Sun distance is 150*10^{6} km? Assume that the orbits
are circular.

Solution:

- Concepts:

Kepler's third law - Reasoning:

T^{2}proportional R^{3}. - Details of the calculation:

R_{J}^{3}= R_{E}^{3}*T_{J}^{2}/T_{E}^{2}= (150*10^{6}km)^{3}*144. R_{J}= 7.86*10^{8}km.

A satellite of mass 200 kg is placed in Earth orbit at a height of 200 km
above the surface. It has a circular orbit. What is the period of
the satellite?

Given: R_{Earth} = 6.4*10^{6 }m, M_{Earth} = 5.98*10^{24 }kg

Solution:

- Concepts:

Motion in a central potential, Kepler's third law - Reasoning:

We are asked to find the period T for a satellite orbiting the Earth in a circular orbit. Since the mass of the Earth (m_{1}) is much greater than the mass of the satellite, we may consider the earth to be stationary. - Details of the calculation:

T^{2 }= (4π^{2}/Gm_{1})R^{3}.

R is the radius of the circular orbit.

R = 6.4*10^{6 }m + 2*10^{5 }m = 6.6*10^{6 }m

R^{3}= 2.9*10^{20 }m^{3}

G = gravitational constant = 6.67*10^{-11 }N-m^{2}/kg^{2}

m_{1}= mass of Earth = 5.98*10^{24 }kg

T = 5.4*10^{3 }s = 90 min.

Io, a satellite of Jupiter, has an orbital period of 1.77 days and an orbital
radius of 4.22*10^{5} km. From these data, determine the mass of
Jupiter.

Solution:

- Concepts:

Motion in a central potential, Kepler's third law - Reasoning:

Two astronomical objects pull each other towards their common center of mass. Each object accelerates towards the center of mass. The acceleration of object 1 is a_{1 }= Gm_{2}/R^{2}, and the acceleration of object 2 is a_{2 }= Gm_{1}/R^{2}. If each object has velocity perpendicular to the direction of its acceleration and v_{1}^{2}/R_{1 }= a_{1}, v_{2}^{2}/R_{2 }= a_{2}, with R_{1}and R_{2}being the distances of object 1 and object 2 from the CM, then both objects orbit their common CM in circular orbits. If object 1 is much more massive than object 2, then the CM lies very close to the center of object 1. Then R_{2}is approximately equal to R and we can write Gm_{1}/R = v_{2}^{2}. If object 2 is in a circular orbit about a much more massive object 1, its speed is given by this formula. We may write v_{2 }= 2πR/T_{2}, where T_{2}is the period of object 2. We then have Gm_{1}/R = (2πR/T_{2})^{2}, or T_{2}^{2 }= (4π^{2}/Gm_{1})R^{3}. - Details of the calculation:

Let object 1 be Jupiter, and object 2 be Io. Then

T_{2}^{2 }= (4π^{2}/Gm_{1})R^{3}.

m_{1 }= (4π^{2}/GT_{2}^{2})R^{3}.^{ }m_{1 }= 4π^{2}(4.22*10^{8}m)^{3}/((6.67*10^{-11}Nm^{2}/kg^{2})(1.77*24*60*60s)^{2}) = 1.9*10^{27}kg.

The source of the first gravitational wave event observed by the LIGO
collaboration in 2015 has been interpreted as the merger of two black holes in a
binary system, each with a mass of roughly 35 solar masses (implying a radius
for the event horizon of about 100 km for each, if assumed spherical), where a
solar mass is 1*.*989*10^{30 }kg. A full understanding
requires general relativity, but assume Newtonian mechanics and Newtonian
gravity as a first approximation for the orbital motion. At the peak
amplitude of the detected gravitational wave, its measured frequency indicated
that the two black holes were revolving around the center of mass about 75 times
per second.

What was the approximate separation of the centers for the
two black holes at this point in the merger event?

Solution**:**

- Concepts:

Kepler's third law, relative motion - Reasoning:

The problem of two interacting particles in their CM frame is equivalent to the problem of a fictitious particle of reduced mass μ moving in a central potential U(r). - Details of the calculation:

Here U(r) = Gm^{2}/r, F(r) = Gm^{2}/r^{2},^{ }r = distance between their centers.

μv^{2}/r = Gm^{2}/r^{2}. v^{2}= Gm^{2}/(μr). μ = m/2. v^{2}= G 2m/r.

(2πr/T)^{2}= G 2m/r. r = [T^{2}G2m/(4π^{2})]^{(1/3)}.

With T = (1/75) s and m = 35*1*.*989*10^{30 }kg we have r = 3.47*10^{5}m = 347 km.

A particle of mass m is released a distance b from a fixed origin of force
that attracts the particle according to the inverse square law F(x) = -k/x^{2}.
Find the time required for the particle to reach the origin. Use this
result to show that, if the Earth were suddenly stopped in its orbit, it would
take approximately 65 days for it to collide with the Sun. Assume
that the Sun is as a fixed point mass and Earth's orbit is circular.

Solution**:**

- Concepts:

Kepler's third law - Reasoning:

For a central force F(r) = -k/r^{2}, the closed orbits are elliptical orbits. All elliptical orbits with the same semi-major axis have the same period. The elliptical orbit with zero angular momentum is a straight line. The motion of a particle in this orbit is in one dimension and with the appropriate orientation of the coordinate system we have F(x) = -k/x^{2}. - Details of the calculation:

The period of the motion of a particle with semi-major axis r can easily be found by considering the circular orbit.

k/r^{2}= mv^{2}/r, (2πr/T)^{2}r = k/m, T^{2}= 4π^{2}mr^{3}/k.

For the particle in the problem r = b/2 and the time t_{0}required to reach the origin is T/2.

t_{0}= ½πb(½mb/k)^{½}.

If the Earth were suddenly stopped in its orbit, its semi-major axis would be halved.

(T_{new}/365 days) = (R_{new}/R)^{3/2}= 2^{-3/2}. T_{new}= 129 days.

t_{0}= T_{new}/2 = 64.5 days.We can also brute-force integrate to find t

_{0}.

From energy conservation: ½mv(x)^{2}= -k/b + k/x = k(b - x)/(xb).

v(x)^{2}= 2k(b - x)/(mxb). dt = dx/v(x).

t_{0}= (½mb/k)^{½}∫_{b}^{0}dx (x/(b - x))^{½}.

let x = v, b - x = u.

∫_{b}^{0}dx√v/√u = √(uv)|_{b}^{0}- ½b∫_{b}^{0}dx/√(vu) = ½b∫_{0}^{b}dx/√(vu) = ½b∫_{0}^{b}dx/√(bx -x^{2}).

Let X = a + bx + cx^{2}with c < 0. Then ∫dx/√X = (-c)^{-½}sin^{-1}((-2cx - b)/(b^{2}- 4ac)^{½}.

∫_{0}^{b}dx/√(bx - x^{2}) = sin^{-1}(1) - sin^{-1}(-1) = π.

t_{0}= ½πb(½mb/k)^{½}.

__Energy and angular momentum conservation__

A comet of mass m approaches the solar system with a
velocity v_{0}, and if it had not been attracted towards the sun, it
would have missed the sun by a distance d. Calculate its minimum distance
z from the sun as it passes through the solar system. Make and state any
reasonable simplifying assumptions.

Solution:

- Concepts:

Conservation of energy and angular momentum - Reasoning:

For the Kepler problem these conservation laws hold.

Key assumptions: Neglect other planets, assume m << M_{sun}, neglect relativity. - Details of the calculation:

At r --> infinity , L = mv_{0}d. L = constant.

At the distance of closest approach r = z, L = mv_{max}z.

mv_{0}d = mv_{max}z, v_{max}= v_{0}d/z.

E = mv_{0}^{2}/2 = constant = mv_{max}^{2}/2 - GMm/z.

Therefore z^{2}+ 2GMz/v_{0}^{2}- d^{2}= 0, z = (G^{2}M^{2}/v_{0}^{4}+ d^{2})^{½}- GM/v_{0}^{2}.

Assume for this problem Earth is a sphere of radius R and mass M. An object of mass m enters Earth's atmosphere at distance R' > R from Earth's center with speed v at angle α from the radial direction. Ignoring any friction or air resistance, at what angle β (from the radial direction) will it hit Earth's surface?

Solution:

- Concepts:

Motion in a central potential, energy and angular momentum conservation - Reasoning:

In a central potential energy and angular momentum are conserved. - Details of the calculation:

At a distance r > R from the center of Earth the object has potential energy -GMm/r.

Energy conservation:

½mv_{R'}^{2}- GmM/R' = ½mv_{R}^{2}- GmM/R,

v_{R}^{2}= v_{R'}^{2}+ 2(R' - R)GM/(RR').

Angular momentum conservation:

mR'v_{R'}sinα = mRv_{R}sinβ, sinβ = sinα (R'/R)(v_{R'}/v_{R}).

sinβ = (R'/R) sinα v_{R'}/(v_{R'}^{2}+ 2(R' - R)GM/(R'R ))^{½}.

Let R' = aR. Then sinβ = a sinα v_{R'}/(v_{R'}^{2}+ 2(a - 1)GM/(aR))^{½}.

We need 0 < sinβ < 1 for the object to hit the Earth. If this condition is not fulfilled than the object reaches its distance of closest approach (sinβ = 1) and continues in its orbit.

A uniform spherical planet of radius a revolves around the sun in
a circular orbit of radius r_{0 }and angular velocity
ω_{0}. It rotates about its axis with
angular velocity Ω_{0} (period T_{0})
normal to the plane of the orbit. Due to tides raised on the planet by the
sun, its angular velocity of rotation is decreasing. Find an expression
which gives the orbital radius r as a function of the angular
velocity Ω of rotation and the parameters r_{0} and T_{0 }at any later or earlier time.

Solution:

- Concepts:

Conservation of angular momentum, Newton's law of gravitation - Reasoning

No external forces act on the sun-planet system, there are no external torques. Therefore the total angular momentum of the system is conserved. Newton's law of gravitation yield a relationship between the period and the orbital radius of circular orbits for planets orbiting a sun (Kepler's third law). - Details of the calculation:

The total angular momentum of the system is the sum of the angular momenta of the revolution and the rotation. Let the z-direction be the direction perpendicular to the plane of the orbit. The direction of all angular momenta is the z-direction.revolution: L

_{rev}= M_{p}r^{2}ω for a spherical orbit.rotation: L

_{rot}= IΩ = (2/5)M_{p}a^{2}Ω.Conservation of angular momentum:

(2/5)M_{p}a^{2}Ω_{0}+ M_{p}r_{0}^{2}ω_{0}= (2/5)M_{p}a^{2}Ω + M_{p}r^{2}ω.^{ }Newton's law of gravitation:

(2/5)a^{2}Ω_{0}+ r_{0}^{2}[GM_{s}/r_{0}^{3}]^{½}= (2/5)a^{2}Ω + r^{2}[GM_{s}/r^{3}]^{½}.^{ }r = (1/GM_{s})[(2/5)a^{2}[(2π/T_{0}) - Ω] + [r_{0}GM_{s}]^{½}]^{2}.

__Kepler orbits__

A space station orbits Earth on a circular trajectory. At some moment the captain decides to change the trajectory by turning on the rocket engine for a very short period of time. During the time the engine was on, it accelerated the station in its direction of motion. As a result, the station speed increased by a factor of α. Provide the conditions, in terms of α, that the new trajectory is elliptic, parabolic or hyperbolic. Justify your answers.

Solution:

- Concepts:

Kepler orbits - Reasoning:

Since the mass of a satellite is negligible compared to the mass of Earth, assume the CM is the center of Earth. For a single object moving in a closed orbit in an attractive 1/r potential, the energy E per unit mass is proportional to inverse of the semi-major axis. - Details of the calculation:

For the Kepler problem we have:E > 0 hyperbola E = 0 parabola E < 0 ellipse The energy per unit mass of the satellite in a circular orbit of radius R is E = -½GM/R, its kinetic energy per unit mass is T = ½GM/R and its potential energy per unit mass is U = -GM/R. E = T + U.

To put it in a parabolic orbit with E = 0, its kinetic energy per unit mass at a distance R from the planet has to double. Its speed therefore has to increase by a factor of √2.

If α > √2 the orbit will be hyperbolic and if 1 < α < √2 the orbit will be elliptic.

For a satellite orbiting a planet, transfer between coplanar circular orbits
can be affected by an elliptic orbit with perigee and apogee distances equal to
the radii of the respective circles as shown in the Figure below. This ellipse
is known as Hohmann transfer orbit.

Assume a satellite is orbiting in a circular orbit of radius r_{p} with
circular orbit speed v_{c}. It is to be transferred into a circular
orbit with radius r_{a}.

(a) Find E_{H}/E_{c}, the ratio of the total energies of the
satellite in the Hohmann and the initial circular orbit.

(b) Determine the equation for the ratio of the speeds v/v_{c }as a
function of the ratio of the distances r/r_{p} from the focus for the
Hohman transfer orbit. Evaluate v/v_{c} at r = r_{p}.

Solution:

- Concepts:

Motion in a central potential of the form -α/r, energy conservation - Reasoning:

All orbits are Kepler orbits. - Details of the calculation:

(a) The acceleration of a satellite in a circular orbit of radius r is a = α/r^{2}= v^{2}/r.

The speed is v = (α/r)^{½}. T = ½mv^{2}= ½αm/r = -½U. E = T + U = ½U = -½mα/r.

We then can write r = αm/(2|E|), |E| = αm/(2r).

This generalizes for an elliptical Kepler orbit to (r_{min}+ r_{max})/2 = αm/(2|E|).

Therefore E_{H}/E_{c}= (2r_{p})/(r_{p}+ r_{a}). Both E_{H}and E_{c}are negative, E_{H}is less negative than E_{c}.

(b) For the Hohmann orbit ,_{ }E_{H}= -αm/(r_{p}+ r_{a}) = ½mv^{2}- αm/r.

½v^{2}= -α/(r_{p}+ r_{a}) + α/r, ½v_{c}^{2}= α/(2r_{p}), (v/v_{c})^{2}= -2r_{p}/(r_{p}+ r_{a}) + 2r_{p}/r.

v/v_{c}= [-2r_{p}/(r_{p}+ r_{a}) + 2r_{p}/r]^{½}.

At r = r_{p}we have v/v_{c}= [-2r_{p}/(r_{p}+ r_{a}) + 2]^{½}= [2r_{a}/(r_{p}+ r_{a})]^{½}.

To put the satellite in the transfer orbit, its speed has to increase at r = r_{p}.

This question is about an elliptical "transfer orbit" from an inner
circular orbit A to an outer circular orbit B. The transfer starts at
point P and is completed at point Q. The transfer orbit is an ellipse
which is tangent to A at point P and tangent to C at point Q.

(a)
Derive a formula for the relationship between v and r for circular orbits.
Is the speed in orbit C greater or less than the speed in orbit A?

(b)
For the transfer, should the satellite speed up or slow down at point P?

(c)
For the transfer, should the satellite speed up or slow down at point Q?

Solution:

- Concepts:

Motion in a potential of the form V(r) = -α/r - Reasoning:

The satellite moves in the gravitational potential of Earth. Since the mass of Earth, M, is much greater than the mass of the satellite, m, we assume that the CM of the system is at the center of Earth. - Details of the calculation:

(a) The acceleration of the satellite in a circular orbit is a = GM/r^{2}= v^{2}/r. The speed is v = (GM/r)^{½}. The speed in orbit C is less than the speed in orbit A.

(b) The transfer orbit is an elliptical orbit. Its semi-major axis is greater that the radius of orbit A and less than the radius of orbit B.

From part (a) we have T = ½mv^{2}= ½GMm/r = -½U. E = T + U = ½U = -½GMm/r.

We then can write r = GMm/(2|E|). This generalizes for an elliptical Kepler orbit to (r_{min}+ r_{max})/2 = GMm/(2|E|).

To increase the semi-major axis we have to decrease |E|, we have to make E less negative. We have to increase the kinetic energy and therefore speed the satellite up at point P.

(c) By the same argument, we again have to increase the speed of the satellite at point Q.

In outer space, two small balls of equal unknown masses m and charges +q and
-q are held at rest a distance d_{0} apart. Then the balls are
simultaneously launched with equal speeds v_{0} in opposite directions
that are perpendicular to the line connecting the balls. During the subsequent
motion of the balls orbit each other, and their minimum speed is v. Find
the mass m of each ball.

Solution:

- Concepts:

The Kepler problem - Reasoning:

The problem of two interacting particles in their CM frame is equivalent to the problem of a fictitious particle of reduced mass μ moving in a central potential U(r).

The motion is in a plane. Energy E and angular momentum**M**are conserved. - Details of the calculation:

Here U(r) = -α/r, where α = q^{2}/(4πε_{0}), and μ = m/2. (Neglect gravity.)

Closed orbits in the given potential are ellipses. The fictitious particle has maximum speed at r_{min}and minimum speed at r_{max}. At these positions dr/dt = 0 and E = M^{2}/(2μr^{2}) - α/r,

The initial conditions specify v_{max}= v_{0}at r_{min}= d_{0}, since we are told that the particle subsequently slows down.

Angular momentum conservation:

M = μd_{0}2v_{0}= md_{0}v_{0}= mr_{max}v.

This is one equation but two unknowns, m and r_{max}.

We need a second equation to eliminate r_{max}.

Energy conservation:

E = ½μ(2v_{0})^{2}- α/d_{0}= mv_{0}^{2}- α/d_{0}= mv^{2}- α/r_{max}= mv^{2}- αv/(v_{0}d_{0}).

m(v^{2}- v_{0}^{2}) = α/d_{0}(v/v_{0}- 1).

m = q^{2}/(4πε_{0}d_{0}v_{0}(v + v_{0})).

The escape speed from the surface of a slowly rotating airless
spherical planet is v_{esc}. What is the minimum initial speed of a
projectile launched from a pole that allows it to land on the equator?

Note:

For Kepler orbits:

p/r = 1 + e cos(φ - φ_{0}),

p = L^{2}/mα, e = (1 + 2EL^{2}/mα^{2})^{½},
L = angular momentum

Solution:

- Concepts:

Kepler orbits - Reasoning:

After the launch, the particle will have an elliptical (Kepler) orbit until it lands again. - Details of the calculation:

For a Kepler orbit p/r = 1 + e cos(φ - φ

_{0}), or r = p/(1 + e cos(φ - φ_{0})).

r_{max}= p/(1 - e), when φ - φ_{0}= π.

In this expression p = L^{2}/mα and e = (1 + 2EL^{2}/mα^{2})^{½}= (1 + 2Ep/α)^{½}.

α = GMm, L = angular momentum, E = energy, M = mass of planet,

m = mass of projectile, m << M.

For the given problem p/R = 1 + (1 + 2Ep/α)^{½}cos(3π/4) = 1 - 2^{-½}(1 + 2Ep/α)^{½}.

(1 - p/R)^{2}= ½ + Ep/α, ½ -2p/R + p^{2}/R^{2}= Ep/α, 1/(2p) - 2/R + p/R^{2}= E/α.

To find the minimum energy we differentiate with respect to p and set the result equal to zero.

-1/(2p^{2}) + R^{2}= 0, p = R/2^{½}.

Therefore E_{min}/α = (√2 - 2)/R.

For the Kepler problem E = ½mv^{2}- α/R.

(√2 - 2)/R = ½mv_{min}^{2}/α - R, (√2 - 1)(2GM/R) = v_{min}^{2}.

The escape speed v_{esc}for the planet is found by setting ½mv_{esc}^{2}= GmM/R, v_{esc }= (2GM/R)^{½}.

v_{min}^{2}= (√2 - 1)v_{esc}^{2}.

A particle of mass m is moving in a central potential of the form V(r) =
-α/r.

(a) What is the total energy of the particle if it is moving in
a circular orbit of radius R? What is its speed?

(b) What is the
energy of the particle if it is moving in an elliptical orbit of semi-major axis
R? What is its speed at r = r_{min}, the perigee of its orbit?

Solution:

- Concepts:

Kepler orbits, U(r) = -mα/r,**F**= -mα/r^{2}(**r**/r) - Reasoning:

The total energy is E = T + U, energy is conserved. - Details of the calculation:

(a) F = mv^{2}/R = mα/R^{2}.

v = (α/R)^{1/2}. T = ½mv^{2}= ½αm/R = -½U. E = T + U = ½U = -½mα/R.

(b) This formula generalizes for an elliptical Kepler orbit to

R = (r_{min}+ r_{max})/2, E = -½αm/((r_{min}+ r_{max})/2) = -½mα/R.

The speed of the particle at r = r_{min}is found from -½mα/R = ½mv^{2}- αm/r_{min}.

v^{2}= -α/(R) + 2α/r_{min}.

Find the maximum time a comet (C) of mass m following a parabolic trajectory around the Sun (S) can spend within the orbit of the Earth (E). Assume that the Earth's orbit is circular and in the same plane as that of the comet.

Solution**:**

- Concepts:

The Kepler problem - Reasoning:

Both the Earth and the comet move in the central potential V(r) = -β/r. Let the radius of the circular orbit of the Earth be a. We need to find the time a comet following a parabolic orbit can spend between r_{min}and a. - Details of the calculation:

The total energy of a particle following a parabolic orbit is zero.

E = ½m_{c}(dr/dt)^{2}+ U_{eff}(r) = 0, U_{eff}(r) = U(r) + M^{2}/(2m_{c}r^{2}) = -m_{c}β/r + M^{2}/(2m_{c}r^{2}).

At r = r_{min}we have dr/dt = 0 and m_{c}β/r_{min}= M^{2}/(2m_{c}r_{min}^{2}).

M^{2}= 2m_{c}^{2}βr_{min}.

We have found the relationship between M and r_{min}.

dr/dt = ((2/m)(E - U_{eff}(r))^{½}.

The time the comet spend between r_{min}and a is

t = 2∫_{rmin}^{a}dr/[(2/m_{c})(m_{c}β/r - M^{2}/(2m_{c}r^{2}))]^{½}.

t = 2∫_{rmin}^{a}dr/[(2/m_{c})(m_{c}β/r - m_{c}βr_{min}/r^{2})]^{½}.

t = (2/β)^{½}∫_{rmin}^{a}rdr/(r - r_{min})^{½}.

∫_{rmin}^{a}rdr/(r - r_{min})^{½}= (2/3)a^{3/2}(1 + 2r_{min}/a)(1 - r_{min}/a)^{½}.

Let x = r_{min}/a.

t = C(1 + 2x)(1 - x)^{½}. Let us find an extremum for t.

dt/dx = 2C(1 - x)^{½}- ½C(1 + 2x)(1 - x)^{-½}= 0.

2(1 - x) - ½(1 + 2x) = 0. 3x = 3/2, x = ½.

Since for r_{min}> a the time t spend within Earth's orbit is zero, the extremum is a maximum.

The comet following a parabolic orbit spends the maximum time between r_{min}and a when r_{min}= a/2.

Then t_{max}= (1/β)^{½}(4/3)a^{3/2}.

Since (a^{3}/β)^{½}= T/2π (Kepler's third law), we have t_{max}= 2T/(3π).

t_{max}= (2*365 days)/(3π) = 77 days.