Classical Mechanics - Conservative Force

Suppose that the gravitational force law between two massive objects were

F₁₂ = r̂₁₂ Gm₁m₂/r₁₂^(2+ɛ) 

where ɛ is a small positive number. Which of the following statements would be FALSE?

1. The total mechanical energy of the planet-Sun system would be conserved.
2. The angular momentum of a single planet moving about the Sun would be conserved.
3. The periods of planets in circular orbits would be proportional to the (3+ɛ)/2 power of their respective orbital radii.
4. A single planet could move in a stationary non circular elliptical orbit about the Sun.
5. A single planet could move in a stationary circular orbit about the Sun.

(GR9677 #23)

Solution:

(A) TRUE.
Gravitational force is a conservative force.
In conservative field, the total mechanical energy is conserved.

(B) TRUE
In conservative field, angular momentum, L is conserved.

(C) TRUE
F_g = F_c
GMm/r^(2+ɛ) = mrω²
GMm/r^(2+ɛ) = mr(2π/T)²
GM/r^(3+ɛ) = 4π²/T²
T² = 4π²r^(3+ɛ)/GM
T ∝ r^(3+ɛ)/2 

(D) FALSE
Central force = centripetal force (F_g = F_c) produces circular orbit.
Non central forces do not produce circular orbit.

(E) TRUE
See (D)

Answer: D

Notes:

Central force:

1. It is a force whose magnitude depends only on the distance between the object and the origin.
2. It is a conservative field, can be expressed as F = − ∇V (the negative gradient of a potential energy).
3. Gravitational force, Coulomb force, and Elastic Force (Harmonic Oscillator) are examples of central (conservative) forces.
4. In conservative field, the net work done by the force is zero, W = ∮_c F ∙ dr = 0 → the total mechanical energy is conserved.
5. Conservative force is irrotional (torque = 0), since curl ∇V or ∇ × ∇V = 0.
6. Torque, τ = dL/dT = 0 → angular momentum, L is conserved
7. Central force = centripetal force (F_g = F_c) produces circular orbit.

Classical Mechanics - Orbital Path

When it is about the same distance from the Sun as is Jupiter, a spacecraft on a mission to the outer planets has a speed that is 1.5 times the speed of Jupiter in its orbit. Which of the following describes the orbit of the spacecraft about the Sun?

A. Spiral
B. Circle
C. Ellipse
D. Parabola
E. Hyperbola

(GR9677 #66)

Solution:

The mission is to outer planets, so the path should not be bounded (no longer a close path): A, B, C are FALSE.

If v_escape = √2 v_circular → Parabola
If v_e  √2 v_c → Hyperbola
Since v_e = 1.5  √2 → Hyperbola

Answer: E

Notes:

To find escape velocity:
F_c = F_g 
mv²/ r = GMm / r² 
v_circular = √(GM / r)

KE = PE_gravity 
½ mv² = GMm / r  
v_escape = √(2GM / r) = √2 √(GM / r)
v_escape = √2 v_circular

Classical Mechanics - Satellite

A Satellite orbits the Earth in a circular orbit. An Astronaut on board perturbs the orbit slightly by briefly firing a control jet aimed toward the Earth’s center. Afterward, which of the following is true of the satellite’s path?

A. It is an ellipse
B. It is a hyperbola
C. It is a circle with larger radius
D. It is a spiral with increasing radius
E. It exhibits many radial oscillations per revolution.

(GR8677 #02)

Solution:

Initially, the object orbits the Earth in a circular orbit.

Perturbs the orbit slightly means giving some extra momentum, so the orbit won't be circular any longer, and will be elliptic, not enough to be Parabolic or Hyperbolic.

Also, logically the astronaut will not want the satellite to have v = v_escape (parabolic) or v v_escape (hyperbolic).

Answer: A


Notes:

Types of Orbits	Eccentricity	Energy	Velocity
Circular	e = 0	E = Vmin
Elliptic	0 e 1	Vmin E 0	v vescape
Parabolic	e = 1	E = 0	v = vescape It will escape the gravitational pull of the planet. If v is increased it will become a hyperbolic orbit.
Hyperbolic	e 1	E 1	v vescape It escapes the gravitational pull of the planet and continues to travel infinitely until it is acted upon by another body with sufficient gravitational force.

The orbital eccentricity, e is the amount by which its orbit deviates from a perfect circle.

Classical Mechanics - Kepler's Law

The period of a hypothetical Earth satellite orbiting at sea level would be 80 minutes. In terms of the Earth’s radius R_e, the radius of a synchronous satellite orbit (period 24 hours) is most nearly

A. 3 R_e
B. 7 R_e
C. 18 R_e
D. 320 R_e
E. 5800 R_e

(GR8677 #75)

Solution:

Kepler’s 3rd Law:

R³ / T²  = constant    

Given:
T_earth = 80 minutes
T_satellite = 24 × 60 minutes

R_e³ / T_e²  = R_s³ / T_s²  

R_s³ = ( T_s² / T_e² ) R_e³ 

R_s³ = (24 × 60 / 80)² R_e³ = (18)² R_e³ 

R_s = (18)^2/3 R_e = (324)^1/3 R_e ≈ 7 R_e

Answer: B