Kepler’s Laws and the Motion of Planets Problems and Solutions

 Problem#1

The center-to-center distance between Earth and Moon is 384 400 km. The Moon completes an orbit in 27.3 days. (a) Determine the Moon’s orbital speed. (b) If gravity  were switched off, the Moon would move along a straight line tangent to its orbit, as described by Newton’s first law. In its actual orbit in 1.00 s, how far does the Moon fall below the tangent line and toward the Earth?

Answer:
Known:
Distance between Earth and Moon, r_Em = 384 400 km
The Moon completes an orbit in 27.3 days or period, T = 27.3 days

(a) the Moon’s orbital speed is
v = 2πr/T
v = 2π(384 400 x 10³ m)/(27.3 x 86400 s) = 1.02 x 10³ m/s = 1.02 km/s

(b) If gravity  were switched off, the Moon would move along a straight line tangent to its orbit, as described by Newton’s first law. In its actual orbit in 1.00 seconds, the Moon falls far below the tangent line and goes to Earth is

y = ½ at² = ½ (v²/r)t²
y = ½ (1.02 x 10³ m/s)² x (1 s)²/(384 400 x 10³ m) = 1.35 mm

The Moon only moves inward 1.35 mm for every 1020 meters it moves along a straight-line path.

Problem#2
Plaskett’s binary system consists of two stars that revolve in a circular orbit about a center of mass midway between them. This means that the masses of the two stars are equal (Fig.1). Assume the orbital speed of each star is 220 km/s and the orbital period of each is 14.4 days. Find the mass M of each star. (For comparison, the mass of our Sun is 1.99 x 10³⁰ kg.)

Fig.1

Answer:
Known:
the orbital speed of each star, v = 220 km/s = 220 x 10³ m/s
the orbital period of each, T = 14.4 days = 14.4 x 86400 s
the mass of our Sun, M_S = 1.99 x 10³⁰ kg
the mass M of each star is applying Newton’s 2nd Law,

∑F = ma,

yields F_g = ma_c for each star:

Gm₁m₂/r₁₂² = m₁v²/r_1c

r_1c is the distance from 1 star to the center of the sun, then

GM²/(2r)² = Mv²/r

We can write r in terms of the period, T, by considering the time and distance of one complete cycle. The  distance traveled in one orbit is the circumference of the stars’ common orbit, so 2πr = vT. Therefore

M = 4v²r/G = [4v²/G][vT/2π] or
M = 2v³T/πG
    = 2(220 x 10³ m/s)²(14.4 x 86400 s)/(π x 6.67 x 10^-11 Nm²/kg²)
M = 1.26 x 10³² kg = 63.3 solar masses

Problem#3
A particle of mass m moves along a straight line with constant speed in the x direction, a distance b from the x axis (Fig. 2). (a) What is the angular momentum (magnitude and direction) of the particle about the origin O? and (b)Show that Kepler’s second law is satisfied by showing that the two shaded triangles in the figure have the same area when t₄ – t₃ = t₂ – t₁.

Fig.2

Answer:
(a) the angular momentum (magnitude and direction) of the particle about the origin O is
L = v x p
L = mv₀b, direction is into paper

(b) consider triangle OO’B

It;s area = v₀t₂b/2

Area of triangle OO’A = v₀t₁b/2

Area of triangle OAB = v₀b(t₂ – t₁)/2

Similarly, area of triangle OCD = v₀b(t₄ – t₃)/2
If time internals equal; namely

t₂ – t₁ = t₄ – t₃
then OAB = OCD in equal intervals of time, hence veriflying Kepler;s 2nd Law!

Problem#4
Io, a moon of Jupiter, has an orbital period of 1.77 days and an orbital radius of 4.22 x 10⁵ km. From these data,  determine the mass of Jupiter.

Answer:
Known:

The orbital period, T = 1.77 d = 1.77 x 86400 s
The orbital radius, r = 4.22 x 10⁵ km

Fig.3

It states that the orbital period, T is related to the distance, r as:
where G is the universal gravitational constant (a very small number for reasons physicists still find hard to explain.)

T² = GMr³/4π²

Rearranging for M should give Jupiter's mass.

M = 4π²r³/GT²
    = 4π² x (4.22 x 10⁵ x 10³ m)³/[6.67 x 10^-11 Nm²/kg² x (1.77 x 86400 s)²]
M = 1.90 x 10²⁷ kg

Problem#4
The Explorer VIII satellite, placed into orbit November 3, 1960, to investigate the ionosphere, had the following orbit parameters: perigee, 459 km; apogee, 2289 km (both distances above the Earth’s surface); period, 112.7 min. Find the ratio v_p/v_a of the speed at perigee to that at apogee.

Answer:
By conservation of angular momentum for the satellite,
r_pv_p = r_av_a
v_p/v_a = [(2289 km + 6.37 x 10³ km)/(459 km + 6.37 x 10³ km)] = 1.27
We do not need to know the period.

Problem#5
Comet Halley (Figure 4) approaches the Sun to within 0.570 AU, and its orbital period is 75.6 years. (AU is the symbol for astronomical unit, where 1 AU = 1.50 x 10¹¹ m is the mean Earth–Sun distance.) How far from the Sun will Halley’s comet travel before it starts its return journey?

Fig.4

Answer:
Known:
The orbital period, T = 75.6 years
By Kepler’s Third Law,

T² = ka³

With a is semi-major axis, then

(75.6 years)² = 1.00[(0.570 AU + y)/2]³
(0.570 AU + y) = 2(75.6 years)^2/3
y = 2(75.6 years)^2/3 – 0.570 AU
y = 35.2 AU

Fig.6