Respuesta :
Answer:
(A) -2.5139 x 10^{10} J, the kinetic energy decreases
(B) 5.0278 x 10^{10} J, the potential energy increases
(C) 2.5139 x 10^{10} J
Explanation:
mass of rocket (Mr) = 5 x 10^{3} kg
initial radius (r) = 7.20 x 10^{6} m
final radius (R) = 8.8 x 10 ^{6} m
mass of the earth (Me) = 5.97 x 10^{24} kg
gravitational constant (G) = 6.67 x 10^{-11} N.m^{2} / kg^{2}
(A) find the change in kinetic energy?
change in kinetic energy = 0.5MrV^{2} - 0.5Mrv^{2}
- where V = final velocity = [tex]\sqrt{\frac{GMe}{R} }[/tex]
V = [tex]\sqrt{\frac{6.67 x 10^{-11} x 5.97 x 10^{24} }{8.8 x 10 ^{6} }[/tex]
V = 6,726.8 m/s
- v = initial velocity = [tex]\sqrt{\frac{GMe}{r} }[/tex]
v = [tex]\sqrt{\frac{6.67 x 10^{-11} x 5.97 x 10^{24} }{7.2 x 10 ^{6} }[/tex]
v = 7,437.76 m/s
change in K.E = (0.5 x 5000 x 6,726.8^{2}) - (0.5 x 5000 x 7,437.76^{2})
change in K.E = -2.5139 x 10^{10} J, the kinetic energy decreases
(B) what is the change in the rockets gravitational potential energy
change in P.E = U2 - U1
U2 - U1 = -([tex]\frac{GMeMr}{R}[/tex]) - (-[tex]\frac{GMeMr}{r}[/tex])
U2 - U1 = -([tex]\frac{6.67 x 10^{-11} x 5.97 x 10^{24} x 5000}{8.8 x 10 ^{6}}[/tex]) - (-[tex]\frac{6.67 x 10^{-11} x 5.97 x 10^{24} x 5000}{7.2 x 10 ^{6}}[/tex])
U2 - U1 = 5.0278 x 10^{10} J, the potential energy increases
(C) what is the work done by the rocket engines
work done = change in K.E + change in P.E
work done = -2.5139 x 10^{10} + 5.0278 x 10^{10} = 2.5139 x 10^{10} J
Answer:
a) the change in kinetic energy will be ΔK = -2.516*10¹⁰ J
b) the change in potential energy will be ΔV = 2*(-ΔK) = 5.032*10¹⁰ J
c) the work required will be W= ΔE= -ΔK = 2.516*10¹⁰ J
Explanation:
since the the rocket is in a stable circular orbit the velocity should be
F gravity = m*a
where F gravity is given by Newton's gravitational law ( if we ignore relativistic effects)
F gravity = M*m*G/R²
since a= radial acceleration , for circular motion:
a=v²/R
then
F gravity = m*a
M*m*G/R²= m*v²/R
thus
M*G/R=v²
M*G/R=v²
for a change in velocity
v₁²= M*G/R₁ and v₂²= M*G/R₂
assuming
mass of the earth M= 5.972 × 10^24 kg
gravitational constant= G= 6.674 * 10⁻¹¹ m³ kg⁻¹ s ⁻²
then
Kinetic energy in 1= K₁ = 1/2* m * v₁² =1/2*m*M*G/R₁ = 1/2* 5.00*10³ kg* 5.972 *10²⁴ kg * 6.674 * 10⁻¹¹ m³ kg⁻¹ s ⁻² / (7.20 * 10⁶m ) = 1.384*10¹¹ J
knowing that
K₁ = 1/2* m * v₁² and K₂ = 1/2* m * v₂²
dividing both equations
K₂/K₁= v₂²/v₁² = R₁/R₂
then
K₂ = K₁ * R₁/R₂ =
the change in kinetic energy will be
ΔK = K₂-K₁ = K₁ * R₁/R₂- K₁ = K₁ *(R₁/R₂-1)
replacing values
ΔK = K₁ *(R₁/R₂-1) = 1.384*10¹¹ J * [ (7.20 * 10⁶m)/ (8.80 * 10⁶m) -1 ] = -2.516*10¹⁰ J
ΔK = -2.516*10¹⁰ J
therefore the kinetic energy decreases
the change in potential energy is
ΔV = M*m*G/R₁ - M*m*G/R₂ = m*v₁² - m*v₂² = 2*(-ΔK) = 2* -(-2.516*10¹⁰ J) = 5.032*10¹⁰ J
ΔV = 2*(-ΔK) = 5.032*10¹⁰ J
therefore the potential energy increases
the work required will be
W= ΔE= ΔK + ΔV = ΔK + 2*(-ΔK) = -ΔK = 2.516*10¹⁰ J
