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Ten days after it was launched toward Mars in December 1998, the Mars Climate Orbiter spacecraft (mass 629 kg) was 2.87×106km from the earth and traveling at 1.20×104km/h relative to the earth.At this time, what were the following. (a) the spacecraft's kinetic energy relative to the earth J (b) the potential energy of the earth-spacecraft system J

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boffeemadrid

Answer:

3494444444.44444 J

-87077491.39453 J

Explanation:

M = Mass of Earth = [tex]6.371\times 10^{6}\ kg[/tex]

G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²

R = Radius of Earth = [tex]6.371\times 10^{6}\ m[/tex]

h = Altitude = [tex]2.87\times 10^9\ m[/tex]

m = Mass of satellite = 629 kg

v = Velocity of spacecraft = [tex]1.2\times 10^4\ km/h[/tex]

The kinetic energy is given by

[tex]K=\frac{1}{2}629\times \left(1.2\times 10^4\times \dfrac{1000}{3600}\right)^2\\\Rightarrow K=3494444444.44444\ J[/tex]

The spacecraft's kinetic energy relative to the earth is 3494444444.44444 J

Potential energy is given by

[tex]U=-\dfrac{GMm}{R_e+h}\\\Rightarrow U=-\dfrac{6.67\times 10^{-11}\times 5.97\times 10^{24}\times 629}{2.87\times 10^9+6.371\times 10^{6}}\\\Rightarrow U=-87077491.39453\ J[/tex]

The potential energy of the earth-spacecraft system is -87077491.39453 J

onyebuchinnaji

The kinetic energy of the spacecraft is 3.49 x 10⁹ J.

The potential energy of the earth-spacecraft system is 1.77 x 10¹³ J.

The given parameters;

  • mass of the spacecraft, m = 629 kg
  • speed of the spacecraft, v = 1.20 x 10⁴ km/h = 3,333.33 m/s
  • height of the spacecraft, h = 2.87 x 10⁶ km = 2.87 x 10⁹ m

The kinetic energy of the spacecraft is calculated as follows;

[tex]K.E = \frac{1}{2} \times m \times v^2\\\\K.E = \frac{1}{2} \times 629 \times (3,333.33)^2\\\\K.E = 3.49 \times 10^{9} \ J[/tex]

The potential energy of the earth-spacecraft system;

P.E = mgh

P.E = (629) x (9.8) x (2.87 x 10⁹)

P.E = 1.77 x 10¹³ J.

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