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AveGali [126]
3 years ago
7

A space transportation vehicle releases a 470-kg communications satellite while in a circular orbit 350 km above the surface of

the Earth. A rocket engine on the satellite boosts it into an orbit 2350 km above the surface of the Earth. How much energy does the engine have to provide for this boost?
Physics
1 answer:
natima [27]3 years ago
8 0

Answer:

E = 3.194 x 10⁹ J = 3.194 GJ

Explanation:

The formula for the absolute potential energy is:

U = - GMm/2r

where,

G = Gravitational Constant = 6.67 x 10⁺¹¹ N m²/kg²

M = mass of Earth = 5.972 x 10²⁴ kg

m = mass of satellite = 470 kg

r = distance between the center of Earth and satellite

Thus, the energy required from engine will be difference between the potential energies.

E = U₂ - U₁

E = - GMm/2r₂ - (- GMm/2r₁)

E = (GMm/2)(1/r₁ - 1/r₂)

where,

r₁ = Radius of Earth + 350 km = 6371 km + 350 km = 6721 km = 6.721 x 10⁶ m

r₂=Radius of Earth + 2350 km=6371 km + 2350 km= 8721 km = 8.721 x 10⁶ m

therefore,

E = [(6.67 x 10⁺¹¹ N m²/kg²)(5.972 x 10²⁴ kg)(470 kg)/2](1/6.721 x 10⁶ m - 1/8.721 x 10⁶ m)

<u>E = 3.194 x 10⁹ J = 3.194 GJ</u>

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Final

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b) This is the reference frame of the center of mass of the system in this case the speed of this frame is the speed of the center of mass

      u = 0.25 m / s

In the direction of movement of the ball

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Initial

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Final

     K_{f}= ½ (m₁ + m₂) v_{f}2

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    ΔK = K₀ -  K_{f}

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These will transform internal system energy

d) In order to find the kinetic energy, we must first find the velocities of the individual in this reference system.

      v₁₀’= v₁₀ -u

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      v_{f} ‘=   v_{f} - u

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Initial

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Final

   k_{f} = ½ (m₁ + m₂) vf’²

  k_{f} = 0

All initial kinetic energy is transformed into internal energy in this reference system

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