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Zero, a hypothetical planet, has a mass of 5.7 x 1023 kg, a radius of 3.0 x 106 m, and no atmosphere. A 10 kg space probe is to be launched vertically from its surface. (a) If the probe is launched with an initial kinetic energy of 5.0 x 107 J, what will be its kinetic energy when it is 4.0 x 106 m from the center of Zero? (b) If the probe is to achieve a maximum distance of 8.0 x 106 m from the center of Zero, with what initial kinetic energy must it be launched from the surface of Zero?

Respuesta :

Answer:

Part a)

[tex]KE = 1.83 \times 10^7 J[/tex]

Part b)

[tex]KE = 7.92 \times 10^7 J[/tex]

Explanation:

Here by energy conservation we can say

initial total energy = final total energy

now we have

[tex]-\frac{Gm_1m_2}{r_1} + \frac{1}{2}mv_1^2 = -\frac{Gm_1m_2}{r_2} + \frac{1}{2}mv_2^2[/tex]

now we have

[tex]-\frac{(6.67\times 10^{-11})(5.7 \times 10^{23})(10)}{3\times 10^6} + 5 \times 10^7 = -\frac{(6.67 \times 10^{-11})(5.7 \times 10^{23})(10)}{4 \times 10^6} + KE[/tex]

now we have

[tex]KE = 1.83 \times 10^7 J[/tex]

Part b)

Now again at maximum height the final kinetic energy will be zero

so we have

[tex]-\frac{Gm_1m_2}{r_1} + \frac{1}{2}mv_1^2 = -\frac{Gm_1m_2}{r_2} + \frac{1}{2}mv_2^2[/tex]

now we have

[tex]-\frac{(6.67\times 10^{-11})(5.7 \times 10^{23})(10)}{3\times 10^6} + KE = -\frac{(6.67 \times 10^{-11})(5.7 \times 10^{23})(10)}{8 \times 10^6} + 0[/tex]

now we have

[tex]KE = 7.92 \times 10^7 J[/tex]

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