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A 500 kg satellite is placed in a circular orbit 6394 km above the surface of the earth. At this elevation, the acceleration of gravity is 4.09 m/s2. Knowing that its orbital speed is 20,000 km/h, determine the kinetic energy of the satellite. (Round the final answer to two decimal places.)

Answer :

Answer:

KE  = 7.7160 GJ

Explanation:

given data

mass = 500 kg

radius = 6394 km

acceleration of gravity = 4.09 m/s²

orbital speed = 20,000 km/h  = 20000 × [tex]\frac{1000}{3600}[/tex]  = 5555.56 m/s

solution

we Kinetic energy is express as

KE = 0.5 × m × v²   .................1

here m is mass and v is velocity

put here value and we get

KE = 0.5 × 500 × 5555.56²

KE = 7716061728.400

KE  = 7.7160 GJ

onyebuchinnaji

The kinetic energy of the satellite at the given orbital speed is [tex]7.72 \times 10^9 \ J[/tex]

The given parameters;

  • mass of the satellite, m = 500 kg
  • orbital speed of the satellite, v = 20,000 km/h

Convert the orbital speed of the satellite to m/s as follows;

[tex]v = 20,000 \ \frac{km}{h} \times \frac{1 \ hour}{3600 \ s} \times \frac{1000\ m}{1 \ km} \\\\v = 5,555.6 \ m/s[/tex]

The kinetic energy of the satellite is calculated as follows;

[tex]K.E = \frac{1}{2} mv^2\\\\K.E = \frac{1}{2} \times (500) \times (5,555.6)^2\\\\K.E = 7.72 \times 10^9 \ J[/tex]

Thus, the kinetic energy of the satellite at the given orbital speed is [tex]7.72 \times 10^9 \ J[/tex]

Learn more about kinetic energy here: https://brainly.com/question/23503524

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