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The Kepler problem

• Nov 05, 2011 -Most geologists believe that the dinosaurs became extinct 65 million years ago when a large comet or asteroid struck the earth, throwing up so much dust that the sun was blocked out for a period of many months. Suppose an asteroid with a diameter of 2.0km and a mass of 1.0×10^13kg hits the earth with an impact speed of 4.0×10^4m/s.
• Orbital Elements at Epoch 2454049.5 (2006-Nov-10.0) TDB Reference: JPL 34 (heliocentric ecliptic J2000).

Kepler's third law

Comets travel around the sun in elliptical orbits with large eccentricities. If a comet has speed 2.1×10^4 when at a distance of 2.0×10^11 from the center of the sun, what is its speed when at a distance of 5.3×1010?

Problem:

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*10⁸ km.

Solution:

• Concepts:
  Kepler's third law
• Reasoning:
  mv²/r = GMm/r². v² = GM/r. (2πr/T)² = GM/r.
  T² = (4π²/(GM))r³, Kepler's third 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 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*10¹¹m 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 (m₁) is much greater than the mass of the comet, we may consider the sun to be stationary.
• Details of the calculation:
  T² = (4π²/Gm₁)R³.
  For elliptical orbits, R denotes the semi-major axis, R = (R_max + R_min)/2.
  R³ = Gm₁T²/4π².
  R³ = (6.67*10^-11 Nm²/kg²)(1.991*10³⁰ kg)(75.6*365*24*60*60 s)²/4π² = 1.91*10³⁷ m³.
  R = 2.67*10¹² m = (R_max + R_min)/2.
  R_max = 5.35*10¹² m - 0.570*1.5*10¹¹ m = 5.26*10¹² m.
  R_max is the maximum distance of the comet from the sun.

Problem:

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*10⁶ km? Assume that the orbits are circular.

Solution:

• Concepts:
  Kepler's third law
• Reasoning:
  T² proportional R³.
• Details of the calculation:
  R_J³ = R_E³*T_J²/T_E² = (150*10⁶ km)³*144. R_J = 7.86*10⁸ km.

Problem:

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*10⁶ m, M_Earth = 5.98*10²⁴ 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 (m₁) is much greater than the mass of the satellite, we may consider the earth to be stationary.
• Details of the calculation:
  T² = (4π²/Gm₁)R³.
  R is the radius of the circular orbit.
  R = 6.4*10⁶ m + 2*10⁵ m = 6.6*10⁶ m
  R³ = 2.9*10²⁰ m³
  G = gravitational constant = 6.67*10^-11 N-m²/kg²
  m₁ = mass of Earth = 5.98*10²⁴ kg
  T = 5.4*10³ s = 90 min.

Problem:

Io, a satellite of Jupiter, has an orbital period of 1.77 days and an orbital radius of 4.22*10⁵ 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 a₁ = Gm₂/R², and the acceleration of object 2 is a₂ = Gm₁/R². If each object has velocity perpendicular to the direction of its acceleration and v₁²/R₁ = a₁, v₂²/R₂ = a₂, with R₁ and R₂ 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 R₂ is approximately equal to R and we can write Gm₁/R = v₂². 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 v₂ = 2πR/T₂, where T₂ is the period of object 2. We then have Gm₁/R = (2πR/T₂)², or T₂² = (4π²/Gm₁)R³.
• Details of the calculation:
  Let object 1 be Jupiter, and object 2 be Io. Then
  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 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*10³⁰ 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) = Gm²/r, F(r) = Gm²/r²,r = distance between their centers.
  μv²/r = Gm²/r². v² = Gm²/(μr). μ = m/2. v² = G 2m/r.
  (2πr/T)² = G 2m/r. r = [T² G2m/(4π²)]^(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 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/x². 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/r², 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/x².
• 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/r² = mv²/r, (2πr/T)²r = k/m, T² = 4π²mr³/k.
  For the particle in the problem r = b/2 and the time t₀ required to reach the origin is T/2.
  t₀ = 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.
  t₀ = T_new/2 = 64.5 days.

  We can also brute-force integrate to find t₀.
  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).
  t₀ = (½mb/k)^½∫_b⁰dx (x/(b - x))^½.
  let x = v, b - x = u.
  ∫_b⁰dx√v/√u = √(uv)|_b⁰ - ½b∫_b⁰dx/√(vu) = ½b∫₀^bdx/√(vu) = ½b∫₀^bdx/√(bx -x²).
  Let X = a + bx + cx² with c < 0. Then ∫dx/√X = (-c)^-½sin^-1((-2cx - b)/(b² - 4ac)^½.
  ∫₀^bdx/√(bx - x²) = sin^-1(1) - sin^-1(-1) = π.
  t₀ = ½πb(½mb/k)^½.

Energy and angular momentum conservation

Problem:

A comet of mass m approaches the solar system with a velocity v₀, 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 = mv₀d. L = constant.
  At the distance of closest approach r = z, L = mv_maxz.
  mv₀d = mv_maxz, v_max = v₀d/z.
  E = mv₀²/2 = constant = mv_max²/2 - GMm/z.
  Therefore z² + 2GMz/v₀² - d² = 0, z = (G²M²/v₀⁴ + d²)^½ - GM/v₀².

Problem:

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, what at 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_R² - GmM/R,
  v_R² = v_R'² + 2(R' - R)GM/(RR').
  Angular momentum conservation:
  mR'v_R'sinα = mRv_Rsinβ, 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.

Problem:

A uniform spherical planet of radius a revolves around the sun in a circular orbit of radius r₀ and angular velocity ω₀. It rotates about its axis with angular velocity Ω₀ (period T₀) 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 r₀ and T₀ 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_pr²ω for a spherical orbit.

  rotation: L_rot = IΩ = (2/5)M_pa²Ω.

  Conservation of angular momentum:
  (2/5)M_pa²Ω₀ + M_pr₀²ω₀ = (2/5)M_pa²Ω + M_pr²ω.
  Newton's law of gravitation:
  (2/5)a²Ω₀ + r₀²[GM_s/r₀³]^½ = (2/5)a²Ω + r²[GM_s/r³]^½.
  r = (1/GM_s)[(2/5)a²[(2π/T₀) - Ω] + [r₀GM_s]^½]².

Kepler orbits

Problem:

• 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 (m₁) is much greater than the mass of the comet, we may consider the sun to be stationary.
• Details of the calculation:
  T² = (4π²/Gm₁)R³.
  For elliptical orbits, R denotes the semi-major axis, R = (R_max + R_min)/2.
  R³ = Gm₁T²/4π².
  R³ = (6.67*10^-11 Nm²/kg²)(1.991*10³⁰ kg)(75.6*365*24*60*60 s)²/4π² = 1.91*10³⁷ m³.
  R = 2.67*10¹² m = (R_max + R_min)/2.
  R_max = 5.35*10¹² m - 0.570*1.5*10¹¹ m = 5.26*10¹² m.
  R_max is the maximum distance of the comet from the sun.

Problem:

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*10⁶ km? Assume that the orbits are circular.

Solution:

• Concepts:
  Kepler's third law
• Reasoning:
  T² proportional R³.
• Details of the calculation:
  R_J³ = R_E³*T_J²/T_E² = (150*10⁶ km)³*144. R_J = 7.86*10⁸ km.

Problem:

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*10⁶ m, M_Earth = 5.98*10²⁴ 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 (m₁) is much greater than the mass of the satellite, we may consider the earth to be stationary.
• Details of the calculation:
  T² = (4π²/Gm₁)R³.
  R is the radius of the circular orbit.
  R = 6.4*10⁶ m + 2*10⁵ m = 6.6*10⁶ m
  R³ = 2.9*10²⁰ m³
  G = gravitational constant = 6.67*10^-11 N-m²/kg²
  m₁ = mass of Earth = 5.98*10²⁴ kg
  T = 5.4*10³ s = 90 min.

Problem:

Io, a satellite of Jupiter, has an orbital period of 1.77 days and an orbital radius of 4.22*10⁵ 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 a₁ = Gm₂/R², and the acceleration of object 2 is a₂ = Gm₁/R². If each object has velocity perpendicular to the direction of its acceleration and v₁²/R₁ = a₁, v₂²/R₂ = a₂, with R₁ and R₂ 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 R₂ is approximately equal to R and we can write Gm₁/R = v₂². 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 v₂ = 2πR/T₂, where T₂ is the period of object 2. We then have Gm₁/R = (2πR/T₂)², or T₂² = (4π²/Gm₁)R³.
• Details of the calculation:
  Let object 1 be Jupiter, and object 2 be Io. Then
  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 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*10³⁰ 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) = Gm²/r, F(r) = Gm²/r²,r = distance between their centers.
  μv²/r = Gm²/r². v² = Gm²/(μr). μ = m/2. v² = G 2m/r.
  (2πr/T)² = G 2m/r. r = [T² G2m/(4π²)]^(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 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/x². 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/r², 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/x².
• 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/r² = mv²/r, (2πr/T)²r = k/m, T² = 4π²mr³/k.
  For the particle in the problem r = b/2 and the time t₀ required to reach the origin is T/2.
  t₀ = 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.
  t₀ = T_new/2 = 64.5 days.

  We can also brute-force integrate to find t₀.
  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).
  t₀ = (½mb/k)^½∫_b⁰dx (x/(b - x))^½.
  let x = v, b - x = u.
  ∫_b⁰dx√v/√u = √(uv)|_b⁰ - ½b∫_b⁰dx/√(vu) = ½b∫₀^bdx/√(vu) = ½b∫₀^bdx/√(bx -x²).
  Let X = a + bx + cx² with c < 0. Then ∫dx/√X = (-c)^-½sin^-1((-2cx - b)/(b² - 4ac)^½.
  ∫₀^bdx/√(bx - x²) = sin^-1(1) - sin^-1(-1) = π.
  t₀ = ½πb(½mb/k)^½.

Energy and angular momentum conservation

Problem:

A comet of mass m approaches the solar system with a velocity v₀, 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 = mv₀d. L = constant.
  At the distance of closest approach r = z, L = mv_maxz.
  mv₀d = mv_maxz, v_max = v₀d/z.
  E = mv₀²/2 = constant = mv_max²/2 - GMm/z.
  Therefore z² + 2GMz/v₀² - d² = 0, z = (G²M²/v₀⁴ + d²)^½ - GM/v₀².

Problem:

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, what at 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_R² - GmM/R,
  v_R² = v_R'² + 2(R' - R)GM/(RR').
  Angular momentum conservation:
  mR'v_R'sinα = mRv_Rsinβ, 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.

Problem:

A uniform spherical planet of radius a revolves around the sun in a circular orbit of radius r₀ and angular velocity ω₀. It rotates about its axis with angular velocity Ω₀ (period T₀) 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 r₀ and T₀ 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_pr²ω for a spherical orbit.

  rotation: L_rot = IΩ = (2/5)M_pa²Ω.

  Conservation of angular momentum:
  (2/5)M_pa²Ω₀ + M_pr₀²ω₀ = (2/5)M_pa²Ω + M_pr²ω.
  Newton's law of gravitation:
  (2/5)a²Ω₀ + r₀²[GM_s/r₀³]^½ = (2/5)a²Ω + r²[GM_s/r³]^½.
  r = (1/GM_s)[(2/5)a²[(2π/T₀) - Ω] + [r₀GM_s]^½]².

Kepler orbits

Problem:

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 Keppler orbits.
• Details of the calculation:
  (a) The acceleration of a satellite in a circular orbit of 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 can write r = αm/(2|E|).
  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) = ½mv² - αm/r.
  ½v² = -α/(r_p + r_a) + α/r, ½v_c² = 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_p/(r_p + r_a)]^½.
  To put the satellite in the transfer orbit, its speed has to increase at r = r_p.

Problem:

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. Wifispoof 3 4 8. 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/r² = v²/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 = ½mv² = 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.

Problem:

In outer space, two small balls of equal unknown masses m and charges +q and -q are held at rest a distance d₀ apart. Then the balls are simultaneously launched with equal speeds v₀ 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 = q²/(4πε₀), 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 = M²/(2μr²) - α/r,
  The initial conditions specify v_max = v₀ at r_min = d₀, since we are told that the particle subsequently slows down.
  Angular momentum conservation:
  M = μd₀2v₀ = md₀v₀ = mr_maxv.
  This is one equation but two unknowns, m and r_max.
  We need a second equation to eliminate r_max.
  Energy conservation:
  E = ½μ(2v₀)² - α/d₀ = mv₀² - α/d₀ = mv² - α/r_max = mv² - αv/(v₀d₀).
  m(v² - v₀²) = α/d₀(v/v₀ - 1).
  m = q²/(4πε₀d₀v₀(v + v₀)).

Problem:

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 - φ₀),
p = L²/mα, e = (1 + 2EL²/mα²)^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 - φ₀), or r = p/(1 + e cos(φ - φ₀)).
  r_max = p/(1 - e), when φ - φ₀ = π.
  In this expression p = L²/mα and e = (1 + 2EL²/mα²)^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 + p²/R² = Ep/α, 1/(2p) - 2/R + p/R² = E/α.
  To find the minimum energy we differentiate with respect to p and set the result equal to zero.
  -1/(2p²) + R² = 0, p = R/2^½.
  Therefore E_min/α = (√2 - 2)/R.
  For the Kepler problem E = ½mv² - α/R.
  (√2 - 2)/R = ½mv_min²/α - R, (√2 - 1)(2GM/R) = v_min².
  The escape speed v_esc for the planet is found by setting ½mv_esc² = GmM/R, v_esc = (2GM/R)^½.
  v_min² = (√2 - 1)v_esc².

Problem:

Polyverse Comet 1 0 0 Mediafire

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?

Polyverse Comet 1 0 0 M I N U S 7

Solution:

• Concepts:
  Kepler orbits, U(r) = -mα/r, F = -mα/r² (r/r)
• Reasoning:
  The total energy is E = T + U, energy is conserved.
• Details of 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 = (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 = ½mv² - αm/r_min.
  v² = -α/(R) + 2α/r_min.

Problem:

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.

1 Equals 0

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) + M²/(2m_cr²) = -m_cβ/r + M²/(2m_cr²).
  At r = r_min we have dr/dt = 0 and m_cβ/r_min = M²/(2m_cr_min²).
  M² = 2m_c²βr_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_cβ/r - M²/(2m_cr²))]^½.
  t = 2∫_rmin^a dr/[(2/m_c)(m_cβ/r - m_cβr_min/r²)]^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 (a³/β)^½ = T/2π (Kepler's third law), we have t_max = 2T/(3π).
  t_max = (2*365 days)/(3π) = 77 days.

Polyverse Comet 1 0 0 Meme