Classical Mechanics - Curved Trajectory

The curvature of Mars is such that its surface drops a vertical distance of 2.0 meters for every 3600 meters tangent to the surface. In addition, the gravitational acceleration near its surface is 0.4 times that near the surface of Earth. What is the speed a golf ball would need to orbit Mars near the surface, ignoring the effects of air resistance?

A. 0.9 km/s
B. 1.8 km/s
C. 3.6 km/s
D. 4.5 km/s
E. 5.4 km/s

(GR9677 #22)

Solution:

y = ½ g_marst²

v_x = x/t 

Given:
y = 2
g_mars = 0.4 g 
x = 3600

2 = ½(0.4)(10)(t)²
t = 1 s
v_x = 3600/1 = 3600 m/s = 3.6 km/s

Answer: C

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.

Quantum Mechanics - Planck Length

The characteristic distance at which quantum gravitational effects are significant, the Planck length can be determined from a suitable combination of the physical constants G, ħ, and c. Which of the following correctly gives the Planck length?

A. Għc
B. Għ²c³
C. G²ħc
D. G^½ħ²c
E. (Għ/c³)^½

(GR9677 #29)

Solution:

Check units:

Let's l = Planck length
 l = G^x ħ^y c^z

G unit = m³ kg⁻¹ s⁻²
ħ unit = Js = Nms = kg m² s⁻¹
c unit = m s⁻¹
l unit = m

m = (m³ kg⁻¹ s⁻²)^x  (kg m² s⁻¹)^y (m s⁻¹)^z

For m:
1 = 3x + 2y + z

For kg:
0 = − x + y
x = y

For s:
0 = − 2x − y − z
0 = − 3x − z

z = − 3x

1 = 3x + 2y + z 

1 = 3x + 2x − 3x = 2x
x = y = 1/2
z = − 3x = − 3/2

l = G^x ħ^y c^z 
= G^1/2 ħ^1/2 c^−3/2
= (Għ/c³)^½

Answer: E

Classical Mechanics - Earth’s gravitational force

Questions 4-5

The magnitude of the Earth’s gravitational force on a point mass is F(r), where r is the distance from the Earth’s center to the point mass. Assume the Earth is a homogeneous sphere of radius R.

What is ? 

A. 32
B. 8
C. 4
D. 2
E. 1

(GR9277 #04)

Solution:

Gravity force: F = GMm/R² → F ∝ 1/R²

F(R) ∝ 1/R²
F(2R) ∝ 1/4R²

F(R) / F(2R) = 4R²/ R² = 4

Answer: C

Classical Mechanics - Gravitation

Questions 4-5

The magnitude of the Earth’s gravitational force on a point mass is F(r), where r is the distance from the Earth’s center to the point mass. Assume the Earth is a homogeneous sphere of radius R. 

Suppose there is a very small shaft in the Earth such that the point mass can be placed at a radius of R/2. What is ?

A. 8
B. 4
C. 2
D. 1/2
E. 1/4

(GR9277 #05)

Solution:

Newton's law of gravity becomes a linear law inside a body → F ∝ r
F(R) ∝ R
F(R/2) ∝ R/2



Answer: C

Classical Mechanics - Gravitation

Which of the following is most nearly the mass of the Earth? (The radius of the Earth is about 6.4 × 10⁶ m)

A. 6 × 10²⁴ kg
B. 6 × 10²⁷ kg
C. 6 × 10³⁰ kg
D. 6 × 10³³ kg
E. 6 × 10³⁶ kg

(GR9277 #19)

Solution:

F =  mg = GMm/r²
M = gr²/G
M = 9.8 × (6.4 × 10⁶)² / (6.67 × 10⁻¹¹)
≈ 10 × 6 × 10¹²⁺¹¹  
= 6 × 10²⁴

Answer: A

Classical Mechanics - 1D Vertical Motion

A rock is thrown vertically upward with initial speed v₀. Assume a friction force proportional to –v, where v is the velocity of the rock, and neglect the buoyant force exerted by air. Which of the following is correct?

1. The acceleration of the rock is always equal to g.
2. The acceleration of the rock is equal to g only at the top of the flight.
3. The acceleration of the rock is always less than g.
4. The speed of the rock upon return to its starting point is v₀.
5. The rock can attain a terminal speed greater than v₀ before it returns to its starting point.

(GR8677 #01)

Solution:

There is a friction force:

• acceleration is not constant → (A) and (C) FALSE
• energy is not conserved so it's initial and final speed is not the same → (D) FALSE
• frictional force slows down the object, so its speed at time t has to be less than its initial speed  → (E) FALSE

Answer: B


Math analysis:

Friction force: F_f  = − kv
The equation of motion with friction force: ma = − mg − kv.

a = − g − (kv/m)
(A) FALSE

At the top of the flight, v = 0
a = − g − (k∙0/m) = − g
(B) TRUE

Moving up: v positive
a = − g − (kv/m) = − g − c
a − g
Moving Down: v negative
a = − g − [k(−v)/m] = − g + c
a − g
(C) FALSE

Classical Mechanics - Uniform Circular Motion

A satellite of mass m orbits a planet of mass M in a circular orbit of radius R. The time required for one revolution is

A. independent of M
B. proportional to √m
C. linear in R
D. proportional to R^3/2
E. proportional to R²

(GR0177 #03)

Solution:

F_c = F_G
mω²R = GmM / R²
ω² = GM / R³
(2π/T)² = GM / R³
T = (4π²R³/ GM)^1/2
T ∝ R^3/2

Answer: D

Classical Mechanics - Satellite

An astronomer observes a very small moon orbiting a planet and measures the moon’s minimum and maximum distances from the planet’s center and the moon’s maximum orbital speed. Which of the following CANNOT be calculated from these measurements?

A. Mass of the moon
B. Mass of the planet
C. Minimum speed of the moon
D. Period of the orbit
E. Semimajor axis of the orbit

(GR0177 #22)

Solution:

Let m = mass of the moon and M = mass of the planet,

F_c = F_G

mv²/r   = GMm/r²

m cancels out so, there’s no way to calculate mass of the moon.

Answer: A

Classical Mechanics - Period of the Motion

A particle of mass m moves in the potential above. The period of the motion when the particle has energy E is

A.

B.

C.

D.

E.

(GR9677 #93)

Solution:

Total Period: T = T_SHO + T_grav

For V = ½kx² → Simple Harmonic Oscillator (SHO)
Period of SHO, T_SHO = 2π√(k/m)

The graph shows only half of the usual SHO potential:
T_SHO = ½ × 2π√(k/m) = π√(k/m)

Total Period: T = π√(k/m) + T_grav

Answer: D


Notes:

To find T_grav with V = mgx:

E = T + V = 0 + mgx 
x = E/mg

Kinematic Equation:

x = v₀t +  ½gt²
v₀ = 0
x = ½gt² = E/mg
t²= 2E/mg²
t = T_grav = √(2E/mg²)

Since the particle has to travel from the origin to the right endpoint and then back to the origin, the total time contribution from this potential:

T_grav = 2√(2E/mg²)

The total period is, T = π√(k/m) + 2√(2E/mg²)

Gravitational and Electric Potential

Identical properties of gravitational and electric potential

Click image to enlarge