Question

A rocket with mass 5.00 X 103 kg is in a circular orbit of radius 7.20 X 106 m around the earth. The rocket’s engines fire for a period of time to increase that radius to 8.80 X 106 m, with the orbit again circular.

(A) What is the change in the rocket’s kinetic energy? Does the kinetic energy increase or decrease?
(B) What is the change in the rocket’s gravitational potential energy? Does the potential energy increase or decrease?
(C) How much work is done by the rocket engines in changing the orbital radius?

Answer 1

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

Step-by-step 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

Answer 2

(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

Step-by-step 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 =
  `\sqrt{(GMe)/(R) }`

V =
`\sqrt{(6.67 x 10^(-11) x 5.97 x 10^(24)  )/(8.8 x 10 ^(6) )`

V = 6,726.8 m/s

• v = initial velocity =
  `\sqrt{(GMe)/(r) }`

v =
`\sqrt{(6.67 x 10^(-11) x 5.97 x 10^(24)  )/(7.2 x 10 ^(6) )`

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 = -(
`(GMeMr)/(R)`) - (-
`(GMeMr)/(r)`)

U2 - U1 = -(
`(6.67 x 10^(-11) x 5.97 x 10^(24) x 5000)/(8.8 x 10 ^(6))`) - (-
`(6.67 x 10^(-11) x 5.97 x 10^(24) x 5000)/(7.2 x 10 ^(6))`)

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