Beings on spherical asteroid have observed that a large rock is approaching their asteroid in a collision course. At 7514 km from the center of the asteroid, the rock has a speed of 136.0 m/s and later at 2823 km it speed has a of 392.0 m/s. Use energy conservation to find the mass of asteroid IT. You can neglect any effects due to the parent star of the asteroid. (G = 6.67 times 10^-11 N. m^2/kg^2)

a. 6.112 times 10^28 kg
b. 4.582 times 10^20 kg
c. 4.582 times 10^21 kg
d. 9.164 times 10^27 kg.

Question

28-12-2019
Physics

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Beings on spherical asteroid have observed that a large rock is approaching their asteroid in a collision course. At 7514 km from the center of the asteroid, the rock has a speed of 136.0 m/s and later at 2823 km it speed has a of 392.0 m/s. Use energy conservation to find the mass of asteroid IT. You can neglect any effects due to the parent star of the asteroid. (G = 6.67 times 10^-11 N. m^2/kg^2)

a. 6.112 times 10^28 kg
b. 4.582 times 10^20 kg
c. 4.582 times 10^21 kg
d. 9.164 times 10^27 kg.

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JemdetNasr JemdetNasr · Answer 1 · 2019-12-28T10:02:25+01:00

Answer:

c. [tex]4.582\times10^{21} kg[/tex]

Explanation:

[tex]r_{i}[/tex] = Initial distance between asteroid and rock = 7514 km = 7514000 m

[tex]r_{f}[/tex] = Final distance between asteroid and rock = 2823 km = 2823000 m

[tex]v_{i}[/tex] = Initial speed of rock = 136 ms⁻¹

[tex]v_{f}[/tex] = Final speed of rock = 392 ms⁻¹

[tex]m[/tex] = mass of the rock

[tex]M[/tex] = mass of the asteroid

Using conservation of energy

Initial Kinetic energy of rock + Initial gravitational potential energy = Final Kinetic energy of rock + Final gravitational potential energy

[tex](0.5) m v_{i}^{2} - \frac{GMm}{r_{i}} = (0.5) m v_{f}^{2} - \frac{GMm}{r_{f}} \\(0.5) v_{i}^{2} - \frac{GM}{r_{i}} = (0.5) v_{f}^{2} - \frac{GM}{r_{f}} \\(0.5) (136)^{2} - \frac{(6.67\times10^{-11}) M}{(7514000)} = (0.5) (392)^{2} - \frac{(6.67\times10^{-11}) M}{(2823000)} \\M = 4.582\times10^{21} kg[/tex]