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

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