Problem:

Use Kepler"s 3rd law to calculate the fixed of the sun, assuming that the orbit the the earth roughly the sun is circular, through radius r = 1.5*108 km.

You are watching: T^2=(4pi^2/gm)r^3

Solution:

Concepts:Kepler"s third lawReasoning:mv2/r = GMm/r2. V2 = GM/r. (2πr/T)2 = GM/r. T2 = (4π2/(GM))r3, Kepler"s 3rd law.Details of the calculation:T = 365 days = 3.15*107 s.M = (4π2/(GT2))r3, M = 2.01*1030 kg.Problem:

Haley"s Comet approaches the sunlight to in ~ 0.570 A.U., and also its orbital duration is 75.6 years. (A.U. Is the abbreviation for expensive units, where 1 A.U. = 1.5*1011m is the typical Earth-Sun distance.) How much from the sun will Haley"s comet travel before it beginning its return journey?

Solution:

Concepts:Motion in a main potential,Kepler"s third lawReasoning:We are asked to uncover Rmax, offered Rmin and the period T because that a comet orbiting the sun. Since the fixed of the sunlight (m1) is much higher than the fixed of the comet, us may consider the sunlight to it is in stationary.Details of the calculation:T2 = (4π2/Gm1)R3.For elliptical orbits, R denotes the semi-major axis, R = (Rmax + Rmin)/2. R3 = Gm1T2/4π2.R3 = (6.67*10-11 Nm2/kg2)(1.991*1030kg)(75.6*365*24*60*60 s)2/4π2 = 1.91*1037m3.R = 2.67*1012 m = (Rmax + Rmin)/2.Rmax = 5.35*1012 m - 0.570*1.5*1011 m = 5.26*1012 m.Rmax is the maximum street of the comet indigenous the sun.Problem:

The time of transformation of earth Jupiter about the sun is TJ ~ 12 years. What is the distance in between Jupiter and the sun if the Earth-Sun distance is 150*106 km? Assume that the orbits space circular.

Solution:

Concepts:Kepler"s 3rd lawReasoning:T2 proportional R3.Details of the calculation:RJ3 = RE3*TJ2/TE2 = (150*106 km)3*144. RJ = 7.86*108 km.Problem:

A satellite of mass 200 kg is placed in planet orbit in ~ a elevation of 200 km over the surface. It has a circular orbit. What is the duration of the satellite?Given: REarth = 6.4*106 m, MEarth = 5.98*1024 kg

Solution:

Concepts:Motion in a central potential,Kepler"s third lawReasoning:We space asked to discover the period T for a satellite orbiting the planet in a circular orbit. Since the mass of the planet (m1) is much better than the mass of the satellite, we may think about the planet to it is in stationary.Details of the calculation:T2 = (4π2/Gm1)R3.R is the radius that the one orbit.R = 6.4*106 m + 2*105m = 6.6*106 m R3 = 2.9*1020 m3G = gravitational constant = 6.67*10-11 N-m2/kg2m1 = fixed of earth = 5.98*1024 kgT =5.4*103s = 90 min.Problem:

Io, a satellite the Jupiter, has actually an orbital period of 1.77 days and also an orbit radiusof 4.22*105 km. From these data, determine the massive of Jupiter.

Solution:

Concepts:Motion in a central potential, Kepler"s third lawReasoning:Two astronomical objects traction each other towards their common center of mass. Every object increases towards the facility of mass. The acceleration of thing 1 is a1 = Gm2/R2, and also the acceleration of object 2 is a2 = Gm1/R2. If each object has actually velocity perpendicular come the direction of its acceleration and v12/R1 = a1, v22/R2= a2, with R1 and also R2 being the distances of object 1 and also object 2 native the CM, climate both objects orbit their common CM in one orbits. If object 1 is much an ext massive than object 2, then the cm lies an extremely close to the facility of object 1. Climate R2 is roughly equal come R and we can write Gm1/R = v22. If thing 2 is in a one orbit about a much more massive object 1, its rate is offered by this formula. We may write v2 = 2πR/T2, whereby T2 is the period of thing 2. Us then have Gm1/R = (2πR/T2)2, or T22 = (4π2/Gm1)R3.Details that the calculation:Let object 1 be Jupiter, and also object 2 it is in Io. Climate T22 = (4π2/Gm1)R3.m1 = (4π2/GT22)R3.m1 = 4π2(4.22*108m)3/((6.67*10-11Nm2/kg2)(1.77*24*60*60s)2) = 1.9*1027kg.Problem:

The resource of the very first gravitational wave occasion observed through the LIGO collaboration in 2015 has actually been interpreted as the closing of 2 black holes in a binary system, each v a massive of around 35 solar masses (implying a radius for the occasion horizon of about 100 km for each, if assumed spherical), where a solar mass is 1.989*1030 kg. A full understanding requires general relativity, but assume Newtonian mechanics and Newtonian gravity as a an initial approximation for the orbital motion. At the peak amplitude of the detect gravitational wave, its measured frequency shown that the 2 black holes to be revolving about the facility of mass about 75 times every second. What to be the almost right separation of the centers because that the 2 black holes at this allude in the merger event?

Solution:

Concepts:Kepler"s 3rd law, family member motionReasoning:The problem of two communicating particles in their CM framework is indistinguishable to the trouble of a fictitious fragment of decreased massμ moving in a main potential U(r). Details of the calculation:Here U(r) = Gm2/r, F(r) = Gm2/r2,r = distance in between their centers.μv2/r = Gm2/r2. V2= Gm2/(μr). μ = m/2. V2 = G 2m/r.(2πr/T)2 = G 2m/r. R = (1/3).With T = (1/75) s and m = 35*1.989*1030 kg we have r = 3.47*105 m = 347 km.Problem:

A fragment of massive m is exit a street b indigenous a fixed beginning of pressure that attractive the fragment according come the train station square legislation F(x) = -k/x2. Find the time required for the particle to with the origin. Usage this an outcome to show that, if the earth were suddenly stopped in the orbit, it would take approximately 65 days for it to collide with the Sun. Assume the the sun is together a fixed point mass and Earth"s orbit is circular.

Solution:

Concepts:Kepler"s third lawReasoning:For a main force F(r) = -k/r2, the closeup of the door orbits space elliptical orbits. Every elliptical orbits with the exact same semi-major axis have the same period. The elliptical orbit v zero angular momentum is a right line. The movement of a fragment in this orbit is in one dimension and also with the proper orientation of the coordinate device we have F(x) = -k/x2.Details of the calculation:The period of the motion of a particle with semi-major axis r can quickly be discovered by considering the circular orbit.k/r2 = mv2/r, (2πr/T)2r = k/m, T2 = 4π2mr3/k.For the bit in the difficulty r = b/2 and also the time t0 forced to reach the origin is T/2.t0 = ½πb(½mb/k)½.If the earth were suddenly quit in that orbit, the semi-major axis would certainly be halved.(Tnew/365 days) = (Rnew/R)3/2 = 2-3/2. Tnew = 129 days.t0 = Tnew/2 = 64.5 days.

We can likewise brute-force incorporate to discover t0.From power conservation: ½mv(x)2 = -k/b + k/x = k(b - x)/(xb).v(x)2 = 2k(b - x)/(mxb). Dt = dx/v(x).t0 = (½mb/k)½∫b0dx (x/(b - x))½.let x = v, b - x = u. ∫b0dx√v/√u = √(uv)|b0 - ½b∫b0dx/√(vu) = ½b∫0bdx/√(vu) = ½b∫0bdx/√(bx -x2).Let X = a + bx + cx2 with c -½sin-1((-2cx - b)/(b2 - 4ac)½.∫0bdx/√(bx - x2) = sin-1(1) - sin-1(-1) = π.t0 = ½πb(½mb/k)½.

Energy and angular inert conservation

Problem:

A comet of massive m approaches the solar mechanism with a velocity v0, and if it had not to be attracted towards the sun, it would have actually missed the sun by a distance d. Calculation its minimum distance z from the sunlight as that passes through the solar system. Make and state any kind of reasonable simplifying assumptions.

Solution:

Concepts:Conservation that energy and also angular momentumReasoning:For the Kepler trouble these conservation regulations hold.Key assumptions: neglect other planets, assume m sun, ignore relativity.Details of the calculation:At r --> infinity , l = mv0d. Together = constant.At the street of closest strategy r = z, l = mvmaxz.mv0d = mvmaxz, vmax = v0d/z.E = mv02/2 = consistent = mvmax2/2 - GMm/z.Therefore z2 + 2GMz/v02 - d2 = 0, z = (G2M2/v04 + d2)½ - GM/v02.Problem:

Assume because that this trouble Earth is a sphere of radius R and also mass M. An object of fixed m beginning Earth"s atmosphere at distance R" > R from Earth"s facility with speed v at edge α from the radial direction. Ignoring any type of friction or waiting resistance, what at angle β (from the radial direction) will certainly it struggle Earth"s surface?

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Solution:

Concepts:Motion in a main potential, energy and angular inert conservationReasoning:In a main potential energy and angular momentum space conserved.Details that the calculation:At a distance r > R from the facility of earth the object has actually potential energy -GMm/r.Energy conservation: ½mvR"2 - GmM/R" = ½mvR2 - GmM/R,vR2 = vR"2 + 2(R" - R)GM/(RR").Angular inert conservation:mR"vR"sinα = mRvRsinβ, sinβ = sinα (R"/R)(vR"/vR).sinβ = (R"/R) sinα vR"/(vR"2 + 2(R" - R)GM/(R"R ))½.Let R" = aR. Then sinβ = a sinα vR"/(vR"2 + 2(a - 1)GM/(aR))½.We need 0 Problem:

A uniform spherical world of radius a revolves about the sun in a circular orbit of radius r0 and angular velocityω0. That rotates about its axis through angular velocity Ω0 (period T0) common to the airplane of the orbit. Due to tides raised on the earth by the sun, that angular velocity that rotation is decreasing. Discover an expression which gives the orbital radius r as a function of the angular velocity Ω that rotation and also the parameters r0 and T0 at any later or earlier time.Solution:

Concepts:Conservation of angular momentum, Newton"s regulation of gravitationReasoningNo external forces action on the sun-planet system, there room no external torques. As such the total angular momentum of the system is conserved. Newton"s regulation of gravitation yield a relationship between the duration and the orbital radius of one orbits because that planets orbiting a sunlight (Kepler"s 3rd law).Details that the calculation:The full angular inert of the system is the amount of the angular momenta of the transformation and the rotation. Allow the z-direction be the direction perpendicular come the aircraft of the orbit. The direction of all angular momenta is the z-direction.

revolution: Lrev = Mpr2ω for a spherical orbit.

rotation: Lrot = IΩ = (2/5)Mpa2Ω.

Conservation the angular momentum:(2/5)Mpa2Ω0 + Mpr02ω0 = (2/5)Mpa2Ω + Mpr2ω.Newton"s regulation of gravitation:(2/5)a2Ω0 + r02½ = (2/5)a2Ω + r2½.r = (1/GMs)<(2/5)a2<(2π/T0) - Ω> + ½>2.Kepler orbits

Problem:

For a satellite orbiting a planet, transfer between coplanar one orbits can be affected by an elliptic orbit v perigee and apogee distances equal come the radii the the corresponding circles as presented in the figure below. This ellipse is known as Hohmann carry orbit.Assume a satellite is orbiting in a one orbit that radius rp v circular orbit speed vc. It is to be transferred right into a circular orbit with radius ra.

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(a) uncover EH/Ec, the ratio of the complete energies that the satellite in the Hohmann and also the initial circular orbit.(b) recognize the equation because that the proportion of the speeds v/vc as a role of the proportion of the distances r/rp from the focus for the Hohman carry orbit. Advice v/vc in ~ r = rp.

Solution:

Concepts:Motion in a main potential that the form -α/r, energy conservationReasoning:All orbits room Keppler orbits.Details the the calculation:(a) The acceleration of a satellite in a one orbit that radius r is a = α/r2 = v2/r. The speed is v = (α/r)½. T = ½mv2 = ½αm/r = -½U. E = T + U = ½U = -½mα/r.We then deserve to write r = αm/(2|E|). This generalizes for an elliptical Kepler orbit to (rmin + rmax)/2 = αm/(2|E|).Therefore EH/Ec = (2rp)/(rp + ra). Both EH and Ec space negative, EH is less an adverse than Ec.(b) because that the Hohmann orbit , EH = -αm/(rp + ra) = ½mv2 - αm/r.½v2 = -α/(rp + ra) + α/r, ½vc2 = α/(2rp), (v/vc)2 = -2rp/(rp + ra) + 2rp/r. V/vc = <-2rp/(rp + ra) + 2rp/r>½.At r = rp we have actually v/vc = <-2rp/(rp + ra) + 2>½ = <2rp/(rp + ra)>½.To put the satellite in the move orbit, the speed needs to increase in ~ r = rp.Problem:

This question is about an elliptical "transfer orbit" indigenous an inside circular orbit A come an outer circular orbit B. The transport starts at suggest P and is perfect at suggest Q. The deliver orbit is an ellipse i beg your pardon is tangent to A at allude P and tangent to C at suggest Q.(a) derive a formula for the relationship in between v and r because that circular orbits. Is the speed in orbit C higher or less than the rate in orbit A?(b) because that the transfer, need to the satellite speed up or sluggish down at allude P?(c) for the transfer, have to the satellite speed up or slow down at allude Q?

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Solution:

Concepts:Motion in a potential of the form V(r) = -α/rReasoning:The satellite move in the gravitational potential that Earth. Because the fixed of Earth, M, is much greater than the fixed of the satellite, m, us assume that the cm of the device is at the center of Earth. Details the the calculation:(a) The acceleration of the satellite in a circular orbit is a = GM/r2 = v2/r. The rate is v = (GM/r)½. The speed in orbit C is much less than the rate in orbit A.(b) The carry orbit is one elliptical orbit. That is semi-major axis is higher that the radius the orbit A and less than the radius the orbit B.From part (a) we have actually T = ½mv2 = ½GMm/r = -½U. E = T + U = ½U = -½GMm/r.We then deserve to write r = GMm/(2|E|). This generalizes because that an elliptical Kepler orbit come (rmin + rmax)/2 = GMm/(2|E|).To increase the semi-major axis we need to decrease |E|, we need to make E less negative. We need to increase the kinetic energy and therefore rate the satellite up at suggest P.(c) by the same argument, we again have to increase the speed of the satellite at allude Q.Problem:

In outer space, two small balls of same unknown masses m and charges +q and -q are hosted at remainder a street d0 apart. Then the balls space simultaneously introduced with same speeds v0 in opposite direction that space perpendicular come the line connecting the balls. During the subsequent movement of the balls orbit each other, and their minimum speed is v. Discover the massive m of every ball.

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Solution:

Concepts:The Kepler problemReasoning:The difficulty of two connecting particles in your CM frame is identical to the difficulty of a fictitious fragment of reduced mass μ relocating in a main potential U(r). The movement is in a plane. Energy E and also angular momentum M
are conserved.Details that the calculation:Here U(r) = -α/r, whereby α = q2/(4πε0), and also μ = m/2. (Neglect gravity.)Closed orbits in the given potential room ellipses. The fictitious particle has maximum speed at rmin and also minimum rate at rmax. At these positions dr/dt = 0 and E = M2/(2μr2) - α/r,The initial problems specify vmax = v0 at rmin = d0, due to the fact that we space told the the bit subsequently slowly down.Angular inert conservation: M = μd02v0 = md0v0 = mrmaxv.This is one equation yet two unknowns, m and rmax. We need a 2nd equation to remove rmax.Energy conservation: E = ½μ(2v0)2 - α/d0 = mv02 - α/d0 = mv2 - α/rmax = mv2 - αv/(v0d0).m(v2 - v02) = α/d0(v/v0 - 1).m = q2/(4πε0d0v0(v + v0)).Problem:

The escape rate from the surface ar of a gradually rotating airless spherical earth is vesc. What is the minimum initial rate of a projectile released from a pole that enables it to land top top the equator?

Note:For Kepler orbits:p/r = 1 + e cos(φ - φ0),p = L2/mα, e = (1 + 2EL2/mα2)½, together = angular momentum

Solution:

Concepts:Kepler orbitsReasoning:After the launch, the fragment will have an elliptical (Kepler) orbit till it lands again.Details the the calculation:
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For a Kepler orbit p/r = 1 + e cos(φ - φ0), or r = p/(1 + e cos(φ - φ0)).rmax = p/(1 - e), once φ - φ0 = π.In this expressionp = L2/mα and e = (1 + 2EL2/mα2)½ = (1 + 2Ep/α)½. α = GMm, l = angular momentum, E = energy, M = fixed of planet, m = massive of projectile, m because that 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 + p2/R2 = Ep/α, 1/(2p) - 2/R + p/R2 = E/α.To find the minimum energy we differentiate with respect to p and collection the result equal come zero.-1/(2p2) + R2 = 0, p = R/2½.Therefore Emin/α = (√2 - 2)/R.For the Kepler problemE = ½mv2 - α/R. (√2 - 2)/R = ½mvmin2/α - R, (√2 - 1)(2GM/R) = vmin2. The escape rate vesc because that the world is uncovered by setup ½mvesc2 = GmM/R, vesc = (2GM/R)½.vmin2 = (√2 - 1)vesc2.

Problem:

A bit of massive m is relocating in a central potential that the type V(r) = -α/r. (a) What is the total energy of the bit if it is moving in a one orbit of radius R? What is the speed?(b) What is the energy of the bit if that is moving in an elliptical orbit that semi-major axis R? What is its rate at r = rmin, the perigee of its orbit?

Solution:

Concepts:Kepler orbits, U(r) = -mα/r, F
= -mα/r2 (r/r)Reasoning:The full energy is E = T + U, energy is conserved.Details that the calculation:(a) F = mv2/R = mα/R2.v = (α/R)1/2. T = ½mv2 = ½αm/R = -½U. E = T + U = ½U = -½mα/R.(b) This formula generalizes for an elliptical Kepler orbit to R = (rmin + rmax)/2, E = -½αm/((rmin + rmax)/2) = -½mα/R.The speed of the particle at r = rmin is uncovered from -½mα/R = ½mv2 - αm/rmin.v2 = -α/(R) + 2α/rmin.Problem:

Find the best time a comet (C) of mass m following a parabolic trajectory about the sunlight (S) can spend in ~ the orbit of the earth (E). Assume the the Earth"s orbit is circular and also in the same airplane as that of the comet.

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Solution:

Concepts:The Kepler problemReasoning:Both the Earth and also the comet relocate in the main potential V(r) = -β/r. Let the radius the the circular orbit the the planet be a. We need to find the time a comet following a parabolic orbit deserve to spend in between rmin and also a.Details of the calculation:The total energy of a particle adhering to a parabolic orbit is zero.E = ½mc(dr/dt)2+ Ueff(r) = 0,Ueff(r) = U(r) + M2/(2mcr2) = -mcβ/r + M2/(2mcr2).At r = rmin we have dr/dt = 0 and also mcβ/rmin = M2/(2mcrmin2).M2 = 2mc2βrmin.We have discovered the relationship between M and also rmin.dr/dt = ((2/m)(E - Ueff(r))½.The time the comet spend in between rmin and a ist = 2∫rmina dr/<(2/mc)(mcβ/r - M2/(2mcr2))>½.t = 2∫rmina dr/<(2/mc)(mcβ/r - mcβrmin/r2)>½.t = (2/β)½∫rmina rdr/(r - rmin)½.∫rmina rdr/(r - rmin)½ = (2/3)a3/2(1 + 2rmin/a)(1 - rmin/a)½.Let x = rmin/a.t = C(1 + 2x)(1 - x)½. Allow us find an extremum because that 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 rmin > a the moment t spend within Earth"s orbit is zero, the extremum is a maximum.The comet complying with a parabolic orbit security the preferably time in between rmin and a as soon as rmin = a/2.Then tmax = (1/β)½(4/3)a3/2.Since (a3/β)½ = T/2π (Kepler"s 3rd law), we have actually tmax = 2T/(3π).tmax = (2*365 days)/(3π) = 77 days.