## Review

#### Problem 1:

A object of mass m orbits a planet of mass M.  The total energy of the object is E.  When its radial distance from the center of the planet is GMm/(8|E|), the radial component of the velocity of the object is zero.  What is the eccentricity e of the orbit?

Hint:  For Kepler orbits 1/r ∝ 1 + e cos(φ - φ0).

Solution:

• Concepts:
Kepler orbits
• Reasoning:
All orbits with the same semi-major axis have the same total energy per unit mass and the same period.
For a circular orbit mv2/R = GMm/R2,  T = ½GMm/R, U = -GMm/R,  E = -½GMm/R, R = ½GMm/|E|.
• Details of the calculation:
The semi-major axis of the orbit of the object is a = ½GMm/|E|.
When R = GMm/(8|E|) = a/4, the radial component of the velocity of the object is zero the objects distance from the center of the planet is Rmin.
Therefore Rmax = 2a - Rmin = 2a - a/4 = 7a/4.
Rmin/Rmax = (1 - e)/(1 + e) = 1/7.
e = -3/4.

#### Problem 2:

A static charge distribution produces a spherically symmetric, radial electric field E  = A exp(-br)/r2(r/r), where A and b > 0 are constants.
(a)  What is the volume charge density ρ(r)?
(b)  What is the total charge?

Solution:

• Concepts:
Gauss' law
• Reasoning:
E = ρ/ε0.  For the given field, the expression is not well defined at r = 0.
We can handle the divergence at r = 0 by using∙(r/r3) = 4πδ(r).
We can also use the integral form of Gauss' law.
• Details of the calculation:
E = A exp(-br)∙(r/r3) + (r/r3)∙(A exp(-br)) = ρ(r)/ε0.
∙(r/r3) = 4πδ(r).
ρ(r) = ρ(r) = 4πε0 A[δ(r) - b exp(-br)/(4πr2)].

We can also use the integral form Gauss' law.
E(r)4πr2 = Qinside0 in SI units.
Qinside = 4πε0 A exp(-br)
ρ(r) = (1/(4πr2))(dQinside /dr) = -ε0 bA exp(-br)/r2, except at r = 0.
As r --> 0 Qinside --> 4πε0 A, there is a point charge at the origin.  It is surrounded by a charge cloud of opposite sign.
Therefore ρ(r) = 4πε0 A[δ(r) -  b exp(-br)/(4πr2)].
(b) As r --> infinite, Qinside --> 0.  The total charge is zero.

#### Problem 3:

The kinetic energy of a non-relativistic mass point equals T = p2/(2m), with p = mv the momentum of the particle.  Find a formally similar expression for the relativistic kinetic energy in terms of the relativistic momentum p = γmv.

Solution:

• Concepts:
Relativistic expression for energy and momentum
• Reasoning:
p = γmv,  E = γmc2 = (m2c4 + p2c2)½.
• Details of the calculation:
γ = (1 - (v/c)2)T = (γ - 1)mc2 = (m2c4 + p2c2)½ - mc2 = mc2(1 + p2/(m2c2))½ - mc2.
This reduces to the non-relativistic expression when p << mc.

#### Problem 4:

Two positive and two negative charges are arranged on the corners of a square whose sides have length L.  Each charge has magnitude Q.  The signs of the charges are as indicated.  Point c is at the center of the square.
(a)  What is the electrostatic potential at point c?
(b)  What is the potential energy associated with the charge at point b due to the other three charges?
(c)  How much work must be done to assemble the charges from ∞?  Clearly explain your reasoning, indicate whether the result is + or -, and explain why.

Solution:

• Concepts:
Electrostatic potential and potential energy
• Reasoning:
We use the principle of superposition.
• Details of the calculation:
(a)  V(r) = k∑qi/|r - ri|, with k = 1/(4πε0).
Put the origin of a coordinate system at the center of the square.
Then r = 0 and |r - ri] = L/√2 for all charges.  Therefore V(c) = 0.
(b)  V(b) = k[-Q/L + Q/L + Q/(√2 L)] = kQ/(√2 L).
Potential energy of a charge -Q at point b:  U = -QV(b) = -kQ2/(√2 L).
(c)  The energy required to assemble a distribution of point charges is
U = k∑all pairs qiqj/|ri - rj|.
U = kQ2(1/L - 1/(√2 L) - 1/L - 1/L - 1/(√2 L) + 1/L) = -kQ2√2/L.
U is negative.

#### Problem 5:

A positron can be made by bombarding a stationary electron with a photon.
γ + e- --> e- + e+ + e-
What is the minimum photon energy?

Solution:

• Concepts:
Relativistic collisions, energy and momentum conservation, frame transformations
• Reasoning:
In relativistic collisions between free particles energy and momentum are always conserved.  Often the physics is best visualized in the center of momentum frame.
• Details of the calculation:
The photon has minimum energy when the reaction products are at rest in the CM frame.
In the lab frame the positron and electrons move together as one particle with mass 3m,  (m = electron mass).
In the CM frame after the collision we have for the length of the momentum 4-vector
P02 - P2 = P02  = (∑p0)2 = (3mc)2 = 9m2c2.
The "length" of the total momentum 4-vector is 9m2c2 before and after the collision in the lab frame.
Before the collision we have
P02 - P2 = (hf/c + mc)2 - (hf/c)2 = m2c2 + 2hfm = 9m2c2.
hf = 4mc2 is the minimum photon energy.

#### Problem 6:

Assume that a star with a radius of 107 km has a purely dipolar magnetic field and that the magnetic field strength in its interior is 100 Gauss.  The star collapses to a neutron star with a radius of 10 km.  Assume that even though mass is lost during this transformation, magnetic flux through an area bounded by the equator is conserved.
(a)  What is the field strength in the interior of the neutron star after the collapse?
(b)  What is the magnetic energy density in the interior of the neutron star?
How does this compare with the energy density from the gravitational field, 2 GeV/fm3 and what does that mean in terms of the ability of the magnetic field to modify the neutron star's structure?

Solution:

• Concepts:
Magnetic flux and energy density
• Reasoning:
Flux:  ΦB = ∫B·dA,  energy density:  uB = Bn2/(2μ0)
• Details of the calculation:
(a)  ΦB = ∫B·dA = (10-2 T)π1020 m2 = Bnπ108 m2.  Bn = 1010 T in the interior of the neutron star.
(b)  The magnetic energy density is  uB = Bn2/(2μ0) = 4*1025 J/m3.
2 GeV/fm3 = 2*1.6*10-10 J/(10-45 m3) = 3.2*1035 J/m3.
The magnetic energy density is a factor of ~10-10 smaller than the gravitational energy density.
The magnetic field cannot modify the structure of the neutron star.