Gravitational Potential Energy Problems and Solutions 2

 Problem#1

Zero, a hypothetical planet, has a mass of 5.0 x 10²³ kg, a radius of 3.0 x 10⁶ m, and no atmosphere. A 10 kg space probe is to be launched vertically from its surface. (a) If the probe is launched with an initial energy of 5.0 x 10⁷ J, what will be its kinetic energy when it is 4.0 x 10⁶ m from the center of Zero? (b) If the probe is to achieve a maximum distance of 8.0 x 10⁶ m from the center of Zero, with what initial kinetic energy must it be launched from the surface of Zero?

Answer:
Mass of planet, M = 5.0 x 10²³ kg
Radius of planet, R = 3.0 x 10⁶ m

(a) If the probe is launched with an initial energy of K_i = 5.0 x 10⁷ J, will be its kinetic energy when it is r_f = 4.0 x 10⁶ m from the center of Zero is by using the concept of energy.

Since the gravitational force is a conservative force, the mechanical energy of the probe-planet system is conserved. The energy conservation equation yields

K_i + U_i = K_f + U_f

K_i + (–GmM/R) = K_f + (–GmM/r_f)
K_f = K_i – GmM[1/R – 1/r_f]

    = (5.0 x 10⁷ J) – (6.67 × 10⁻¹¹ Nm²/kg²)(10 kg)(5.0 x 10²³ kg)[{1/3.0 x 10⁶ m) – (1/4.0 x 10⁶ m)]
K_f = 2.22 x 10⁷ J = 22.2 MJ

(b) When the probe reaches the maximum height, its speed becomes zero. The energy conservation equation leads to

K_i + U_i = K_f + U_f
K_i + (–GmM/R) = 0 + (–GmM/r_f)
K_f = GmM[1/R – 1/r_f]
    = (6.67 × 10⁻¹¹ Nm²/kg²)(10 kg)(5.0 x 10²³ kg)[{1/3.0 x 10⁶ m) – (1/4.0 x 10⁶ m)]
K_f = 2.779 X 10⁷ J = 27.79 MJ

Problem#2
In deep space, sphere A of mass 20 kg is located at the origin of an x axis and sphere B of mass 10 kg is located on the axis at x = 0.80 m. Sphere B is released from rest while sphere A is held at the origin. (a) What is the gravitational potential energy of the twosphere system just as B is released? (b) What is the kinetic energy of B when it has moved 0.20 m toward A?

Answer:
Known:
Mass of sphere A, m_A = 20 kg
Mass of sphere B, m_B= 10 kg
Position A, x_A = 0    
Position B, x_B = 0.80 m
Position B, x_B’ = 0.80 m ─ 0.20 m = 0.60 m

(a) the gravitational potential energy of the twosphere system just as B is released is

U = ─Gm_Am_B/x_B
U = ─(6.67 × 10⁻¹¹ Nm²/kg²)(20 kg)(10 kg)/(0.8 m) = ─1.67 x 10⁻⁸ J

(b) the kinetic energy of B when it has moved 0.20 m toward A is

Conservation of energy given by

U_i + K_i = U_f + K_f         
with K_i = 0 and
U_f = ─Gm_Am_B/x’_B = ─(6.67 × 10⁻¹¹ Nm²/kg²)(20 kg)(10 kg)/(0.6 m) = ─2.23 x 10⁻⁸ J
then
U_i + K_i = U_f + K_f
─1.67 x 10⁻⁸ J + 0 = ─2.23 x 10⁻⁸ J + K_f
K_f = 5.6 x 10⁻⁹ J

Problem#3
(a) What is the escape speed on a spherical asteroid whose radius is 500 km and whose gravitational acceleration at the surface is 3.0 m/s²? (b) How far from the surface will a particle go if it leaves the asteroid’s surface with a radial speed of 1000 m/s? (c) With what speed will an object hit the asteroid if it is dropped from 1000 km above the surface?

Answer:
(a) the particle just escape if its kinetic energy is zero when it is infinitely far from the asteroid. The final potential and kinetic energies are both zero.
Conservation of energy yields

U_i + K_i = 0
With Initially the particle is at the surface of the asteroid and has potential energy

U_i = ─GMm/R, then
─GMm/R + ½ mv² = 0
v² = 2GM/R = 2{GM/R²}R = 2aR
v² = 2 x 3.0 m/s² x 500 x 10³ m
v = 1.7 x 10³ m/s

(b) the final potential energy is U_f = ─GMm/(R + h) and the final energy kinetic is K_f = 0,
 Conservation of energy yields

U_i + K_i = U_f
─GMm/R + ½ mv² = ─GMm/(R + h)
─{GM/R²}R + ½ v² = ─{GM/R²}{R²/(R + h)}
─aR + ½ v² = ─aR²/(R + h)







(c) initially the particle is a distance h above the surface and is at rest. Its potential energy is U_i = ─GMm/(R + h) and the initial energy kinetic is K_i = 0, the final potential energy is U_f = ─GMm/R,

 Conservation of energy yields

U_i + K_i = U_f
─GMm/(R + h) = ─GMm/R + ½ mv_f²
─{GM/R²}{R²/(R + h)} = ─{GM/R²}R + ½ v²
─aR²/(R + h) = ─aR + ½ v²

Problem#4

Fig.1

Figure 1a shows four particles, each of mass 20.0 g, that form a square with an edge length of d = 0.600 m. If d is reduced to 0.200 m, what is the change in the gravitational potential energy of the four-particle system?

Answer:
Consider four masses each of mass m at the corners of square of side d. We have four mass pairs at distance d and two diagonal pairs at distance d√2.

Hence,

U = ─4Gm²/d ─ 2Gm²/(d√2)   (this becomes the initial gravitational potential energy)
U =  ─2Gm²{2/d + 1/(d√2)}
U = ─2(6.67 × 10⁻¹¹ Nm²/kg²)(0.02 kg)²{2/0.600 m + 1/0.600√2 m}
U = ─2.41 x 10⁻¹³ J

If d is reduced to d’, then
U’ = ─4Gm²/d’ ─ 2Gm²/(d’√2)
U’ =  ─2Gm²{2/d’ + 1/(d’√2)}
U’ = ─2(6.67 × 10⁻¹¹ Nm²/kg²)(0.02 kg)²{2/0.200 m + 1/0.200√2 m}
U’ = ─7.22 x 10⁻¹³ J

the change in the gravitational potential energy of the four-particle system is
ΔU = U’ ─ U = ─4.81 x 10⁻¹³ J

Share :