Kepler’s Laws and the Motion of Planets Problems and Solutions 2

 Problem#1

Two planets X and Y travel counterclockwise in circular orbits about a star as in Figure 1. The radii of their orbits are in the ratio 3:1. At some time, they are aligned as in Figure 1a, making a straight line with the star. During the next five years, the angular displacement of planet X is 90.0°, as in Figure 1b. Where is planet Y at this time?

Fig.1

Answer:

the mass M of each star is applying Newton’s 2nd Law,

∑F = ma,

yields F_g = ma_c for each star:

Gm_planetM/r² = m_planetv²/r
GM/r = v²
With v = ωr, then
GM/r = ω²r²
GM = ω²r³

Therefore, the relationship between ω_x and ω_y is
ω_y = ω_x[r_x/r_y]^3/2
     = (90⁰/5 yr)(3)^3/2
ω_y = 468⁰/5 yr

so, planet Y has turned through 1.30 revolutions.

Problem#2
A synchronous satellite, which always remains above the same point on a planet’s equator, is put in orbit around Jupiter to study the famous red spot. Jupiter rotates about its axis once every 9.84 h. Use the data of Table to find the altitude of the satellite.

Answer:
Centripetal Force:
F_C = M_Sv²/r
    = M_S[2π (R_J + h)/T]²/(R_J + h)
F_C = 4π² M_S (R_J + h)/T²

with h is the position of the satellite from the surface of Jupiter
Gravitational Force:

F_g = GM_SM_J/r² = GM_SM_J/(R_J + h)²

Equating the two forces:

4π² M_S (R_J + h)/T² = GM_SM_J/(R_J + h)²
(R_J + h)³ = GM_JT²/(4π²)
(R_J + h)³ = (6.673 x 10^-11 Nm²/kg²)(1.90 x 10²⁷ kg)(9.84 x 3600 s)²/(4π²) = 4.03 x 10²⁴ m³
h = 8.92 x 10⁷ m above the planet

Problem#3
Neutron stars are extremely dense objects that are formed from the remnants of supernova explosions. Many rotate very rapidly. Suppose that the mass of a certain spherical neutron star is twice the mass of the Sun and its radius is 10.0 km. Determine the greatest possible angular speed it can have so that the matter at the surface of the star on its equator is just held in orbit by the gravitational force.

Answer:
Centripetal Force:
F_C = mv²/R_S = mR_Sω²
Gravitational Force:
F_g = GM_Sm/R_S²

The gravitational force on a small parcel of material at the star’s equator supplies the necessary
centripetal force:

mR_Sω² = GM_Sm/R_S²

So

ω² = GM_S/R_S³

ω² = (6.673 x 10^-11 Nm²/kg²)(2 x 1.991 x 10³⁰ kg)/(10.0 x 10³ m)³
ω = 1.63 x 10⁴ rad/s

Problem#4
Suppose the Sun’s gravity were switched off. The planets would leave their nearly circular orbits and fly away in straight lines, as described by Newton’s first law. Would Mercury ever be farther from the Sun than Pluto? If so, find how long it would take for Mercury to achieve this passage. If not, give a convincing argument that Pluto is always farther from the Sun.

Answer:

Centripetal Force:

F_C = M_Pv²/r

Gravitational Force:

F_g = GM_SM_P/r²

Equating the two forces:

M_Pv²/r = GM_SM_P/r²
v² = GM_S/r

For Mercury:
v² = GM_S/r = (6.673 x 10^-11 Nm²/kg²)(1.991 x 10³⁰ kg)/(5.79 x 10¹⁰ m) = 2.29 x 10⁹  m²/s²
v = 47,902 m/s

For Pluto:
v² = GM_S/r = (6.673 x 10^-11 Nm²/kg²)(1.991 x 10³⁰ kg)/(5.91 x 10¹² m) = 2.25 x 10⁷
v = 4,741 m/s

With greater speed, Mercury will ultimately move further from the Sun than Pluto.

(Mercury distance from sun)² = d_M0² + (v_Mt)²
(Pluto distance from sun)² = d_P0² + (v_Pt)²

So that

d_M0² + (v_Mt)² = d_P0² + (v_Pt)²
d_P0² - d_M0² =  (v_M² - v_P² )t²

t² = [(5.91 x 10¹² m)² - (5.79 x 10¹⁰ m)²]/[(229.463 – 2.24804) x 10⁷ m²/s²]
t = 1.239789 x 10⁸ s
t = 34,439 hr = 1435 days = 3.93 years

Problem#5
As thermonuclear fusion proceeds in its core, the Sun loses mass at a rate of 3.64 x 10⁹ kg/s. During the 5 000-yr period of recorded history, by how much has the length of the year changed due to the  loss of mass from the Sun? Suggestions: Assume the Earth’s orbit is circular. No external torque acts on the Earth–Sun system, so its angular momentum is  conserved. If x is small compared to 1, then (1 + x)ⁿ is nearly equal to 1 + nx.

Answer:
Centripetal Force:

F_C = mv²/r

Gravitational Force:

F_g = GM_Sm/r²

For the Earth,

∑ F = ma
GM_Sm/r² = mv²/r
GM_S/r³ = (2π/T)²
GM_ST² = 4π²r³

Also the angular momentum

L = mvr = m r(2πr/T) is a constant for the Earth

We eliminate

r² = RT/2πm

between the equations:

GM_ST² = 4π²(RT/2πm)^3/2
GM_ST^1/2 = 4π²(R/2πm)^3/2

Now the rate of change is described by

GM_S  x (½T^-1/2)dT/dt + G(dM_S/dt x T^1/2) = 0

then

dT/dt = –(dM_S/dt)(2t/M_S) ≈ ΔT/T
ΔT ≈ –Δt(dM_S/dt)(2t/M_S)
ΔT = –5000 yr(3.16 x 10⁷ s/1 yr)(–3.64 x 10⁹ kg/s)(2 x 1 yr/ 1.991 x 10³⁰ kg)
ΔT = 1.82 x 10^-2 s  

Share :