Gravitational Potential Energy Problems and Solutions

  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, m_lo = 8.94 x 10²² kg
Radius of Io, r_lo = 1815 km = 1.815 x 10⁶ m

Fig.1

Can use the speed of the material on Io to solve for the height it woold reack on Earth. Need Io’s gravity, first,

g_Io = GM_Io/r_Io²
g_Io = (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:

v_f² = v_i² + 2a(y – y₀)
with v_f  = 0, then
0 = v_i² + 2(–1.796 m/s²)(5,0 x 10⁵ m)
V_i = 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 = –GM_Em_orbiter/r_orbiter
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:

g_E = GM_E/R_E² and g_M = GM_M/R_M²

For comparison, the ratio between g_M and g_E is taken

g_M/g_E = [M_M­/M_E][R_E/R_M]²
g_M/(9.81 m.s^-2) = [0.11][1.3 x 10⁴ km/6.9 x 10³ km]²
g_M = 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

ρ_M/ρ_E = [M_M­/M_E][V_E/V_M]

with V is the volume of the planet, V = 4πR³/3, then

ρ_M/ρ_E = [M_M­/M_E]R_E/R_M]³
ρ_M/ρ_E = [0.11][1.3 x 10⁴ km/6.9 x 10³ km]³
ρ_M/ρ_E = 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 R_E 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

We're given that: v₀ = 0.500v_esc

Using energy conservation, we have

½ mv₀² ─ GmM_E/R_E = ─ GmM_E/r
½ m(GM_E/2R_E) ─ GmM_E/R_E = ─ GmM_E/r
3/4R_E = 1/r

Therefore,

r = 4R_E/3

(a) Conservation of mechanical energy gives:

K_i + U_i = K_f + U_f

Let's say the projectile reaches a radial distance r before it stops.
We're given that:

K_i = 0.500 × K_esc
K_i = 0.500 × ½ mv_esc² = 0.500 × ½ m x 2GM/R_E

Now, the final kinetic energy of the projectile is zero. So, we have:

K_i + U_i = U_f
0.500 × ½ m x 2GM/R_E ─ GmM_E/R_E = ─ GmM_E/r
1/2R_E ─ 1/R_E = ─ 1/r

or,
r = 2R_E

(c) Conservation of mechanical energy gives:

K_i + U_i = K_f + U_f

Now, the projectile will just escape when K_i + U_i = 0

Therefore, least initial mechanical energy of the projectile is zero.