A994-kg satellite orbits the earth at a constant altitude of 99-km. (a) how much energy must be added to the system to move the satellite into a circular orbit with altitude 210 km? 1033 incorrect: your answer is incorrect. how is the total energy of an object in circular orbit related to the potential energy? mj (b) what is the change in the system's kinetic energy? mj (c) what is the change in the system's potential energy?

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a)   3.11 10²⁴ MJ , b)3.11  10²⁴ MJ,  c) 1.03 10³ MJ

Explanation:

Let's look for the potential energy of the satellite based on its height,

      F = -dU / dr

     dU = - F dr

We integrate and make zero energy for infinite distance

      U = G m M int dr / r²

      U = G m M (-1 / r)

      U = - G m M (1 / r)

With this expression we can calculate the energy that the satellite has for the initial height and subtract it from the energy that the final height, this difference is the energy that we must supply to the system

           ΔU = - G m M (1 / rf - 1 / ri)

Let's look for the distance that is from the center of the Earth

           r = Re + h

           r = 6.37 10⁶ + 210 10³

          r = 6.58 10⁶ m

           r₀ = 6.37 10⁶ + 99 10³

           r₀ = 6,469 10⁶ m

         ΔU = 6.67 10⁻¹¹ 994 5.98 10²⁴ (1 / 6.58 10⁶ - 1 / 6.469 10⁶)

         ΔU = 3.9647 10¹⁷ (0.152 10⁻⁶ - 0.1546 10⁻⁶)

        ΔU = 1.03 10⁹ J

a) the amount of energy to change the orbit satellite is the potential energy plus the kinetic energy

       ΔEm = DK + DU

       ΔEm = 1.03 10⁹ J + 3.11 10³⁰ J

       ΔEm = 3.11 10³⁰ J

      ΔEm = 3.11 10²⁴ MJ

Total energy is the sum of the kinetic energy of rotation plus potential energy

b) To calculate the change in kinetic energy, let's use Newton's second law

       F = m a

       a = v² / r = (wr)² / r = w² r

       G m M / r² = m w² r

       w² = G M / r³

The change in kinetic energy is

        K = ½ I w²

        I = m R²

Let's replace

       K = ½ (m r²) G m / r³

       K = ½ m² G / r

      ΔK = ½ m² G (1 / r - 1 / r₀)

      ΔK = ½ (5.98 10²⁴)² 6.67 10⁻¹¹ (1 / 6.58 10⁶ m - 1 / 6.469 10⁶)

      ΔK = 119.26 10³⁷ (10⁻⁶ 2.61 10⁻³)

     ΔK = 311.27 10²⁸ J

     ΔK = 3.11 10³⁰ J

     ΔK = 3.11  10²⁴ MJ

c) We calculate the change in potential energy

      ΔU = 1.03 10⁹ J

      ΔU = 1.03 10³ MJ

hi there!

your question asks at which segment does freezing occur.

since we're talking about water, the freezing point of water will be 32 degrees fahrenheit or below.

hope this ! ~

multiply 2kg by latent heat of vap