Problem#1
Volcanoes on Io. Jupiter’s moon Io has active volcanoes (in fact, it is the most volcanically active body in the solar system) that eject material as high as 500 km (or even higher) above the surface. Io has a mass of 8.94 x 1022 kg and a radius of 1.815 km. Ignore any variation in gravity over the 500-km range of the debris. How high would this material go on earth if it were ejected with the same speed as on Io?Answer:
Known:
Mass of Io, mlo = 8.94 x 1022 kg
Radius of Io, rlo = 1815 km = 1.815 x 106 m
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| Fig.1 |
gIo = GMIo/rIo2
gIo = (6.67 × 10−11 Nm2/kg2)(8.94 x 1022 kg)/(1.815 x 106 m)2 = 1.796 m/s2
with this speed, can now solve for the speed of the material on Io using the following kinematic equation:
vf2 = vi2 + 2a(y – y0)
with vf = 0, then
0 = vi2 + 2(–1.796 m/s2)(5,0 x 105 m)
Vi = 1.34 m/s
Problem#2
Ten days after it was launched toward Mars in December 1998, the Mars Climate Orbiter spacecraft (mass 629 kg) was 2.87 x 106 km from the earth and traveling at 1.20 x 104 km/h relative to the earth. At this time, what were (a) the spacecraft’s kinetic energy relative to the earth and (b) the potential energy of the earth–spacecraft system?
Answer:
(a) the spacecraft’s kinetic energy relative to the earth is
K = ½ mv2
With v = 1.20 x 104 km/h = 3.33 x 103 m/s
K = ½ mv2 = ½ x 629 kg x (3.33 x 103 m/s)2 = 3.49 x 109 J = 3.49 GJ
(b) the potential energy of the earth–spacecraft system is
U = –GMEmorbiter/rorbiter
U = –(6.67 × 10−11 Nm2/kg2)(5.97 × 1024 kg)(629 kg)/(2.87 x 109 m)
U = –8.73 x 107 J = –87.3 MJ
Problem#3
The mean diameters of Mars and Earth are 6.9 x 103 km and 1.3 x 104 km, respectively. The mass of Mars is 0.11 times Earth’s mass. (a) What is the ratio of the mean density (mass per unit volume) of Mars to that of Earth? (b)What is the value of the gravitational acceleration on Mars? (c) What is the escape speed on Mars?
Answer:
(a) The surface gravitational acceleration of a planet is given by the relation (note that according to Newton’s shell theorem, the planet attracts a particle that is at the surface with all the planet’s mass, acting as if it is concentrated at the planet’s centre)
g = GM/R2
where M and m are the masses of the planet and particle and R is the radius of the planet. Thus, we have for the Earth and Mars:
gE = GME/RE2 and gM = GMM/RM2
For comparison, the ratio between gM and gE is taken
gM/gE = [MM/ME][RE/RM]2
gM/(9.81 m.s-2) = [0.11][1.3 x 104 km/6.9 x 103 km]2
gM = 9.81 m.s-2 x 0.39 = 3.82 m.s-2
(a) the ratio of the mean density (mass per unit volume) of Mars to that of Earth is
ρM/ρE = [MM/ME][VE/VM]
with V is the volume of the planet, V = 4πR3/3, then
ρM/ρE = [MM/ME]RE/RM]3
ρM/ρE = [0.11][1.3 x 104 km/6.9 x 103 km]3
ρM/ρE = 7.3 x 10-4
(c) the escape speed on Mars is
Problem#4
A projectile is shot directly away from Earth’s surface. Neglect the rotation of Earth.What multiple of Earth’s radius RE gives the radial distance a projectile reaches if (a) its initial speed is 0.500 of the escape speed from Earth and (b) its initial kinetic energy is 0.500 of the kinetic energy required to escape Earth? (c) What is the least initial mechanical energy required at launch if the projectile is to escape Earth?
Answer:
(a) the escape speed from Earth given by
Using energy conservation, we have
½ mv02 ─ GmME/RE = ─ GmME/r
½ m(GME/2RE) ─ GmME/RE = ─ GmME/r
3/4RE = 1/r
Therefore,
r = 4RE/3
(a) Conservation of mechanical energy gives:
Ki + Ui = Kf + Uf
Let's say the projectile reaches a radial distance r before it stops.
We're given that:
Ki = 0.500 × Kesc
Ki = 0.500 × ½ mvesc2 = 0.500 × ½ m x 2GM/RE
Now, the final kinetic energy of the projectile is zero. So, we have:
Ki + Ui = Uf
0.500 × ½ m x 2GM/RE ─ GmME/RE = ─ GmME/r
1/2RE ─ 1/RE = ─ 1/r
or,
r = 2RE
(c) Conservation of mechanical energy gives:
Ki + Ui = Kf + Uf
Now, the projectile will just escape when Ki + Ui = 0
Therefore, least initial mechanical energy of the projectile is zero.
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