Your starship lands on a mysterious planet. As chief scientist-engineer, you make the following measurements: A 2.50-kg stone thrown upward from the ground at 12.0 m/s returns to the ground in 6.00 s; the circumference of the mysterious planetat the equator is 2 ×10Okm; and there is no appreciable atmosphere. The starship commander asks for the following information: (a) What is the mass of this mysterious planet? (b) If the starship goes into a circular orbit 30,000 km above the surface of the planet, how many hours will it take the ship to complete one orbit?

Question

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Respuesta :

aristocles aristocles · Answer 1 · 2019-08-19T03:12:09+02:00

Answer:

Part a)

[tex]M = 6.08 \times 10^{19} kg[/tex]

Part b)

[tex]T = 4510 hours[/tex]

Explanation:

As we know that stone is thrown upwards with speed

[tex]v_i = 12 m/s[/tex]

Now it returns back to the surface of Earth after t = 6 s

so the displacement of the stone is zero

[tex]\Delta y = 0 = v t + \frac{1}{2}at^2[/tex]

[tex]0 = 12 t - \frac{1}{2}g t^2[/tex]

[tex]g = \frac{2(12)}{t}[/tex]

[tex]g = 4 m/s^2[/tex]

Part a)

Now we know that the circumference of the planet at the equator is of length

[tex]L = 2 \times 100 km[/tex]

[tex]2\pi R = 2\times 10^5 m[/tex]

[tex]R = 3.2 \times 10^4 m[/tex]

Now we have formula of acceleration due to gravity as

[tex]g = \frac{GM}{R^2}[/tex]

[tex]4 = \frac{6.67 \times 10^{-11} M}{(3.2 \times 10^4)^2}[/tex]

[tex]M = 6.08 \times 10^{19} kg[/tex]

Part b)

Time to complete one revolution around the planet is given as

[tex]T = 2\pi\sqrt{\frac{r^3}{GM}}[/tex]

here we know that

r = distance from center of the planet

[tex]r = 3.2 \times 10^4 + 3\times 10^7 = 3.003 \times 10^7 m[/tex]

now we have

[tex]T = 2\pi\sqrt{\frac{(3.003\times 10^7)^3}{(6.67 \times 10^{-11})(6.08\times 10^{19})}}[/tex]

[tex]T = 4510 hours[/tex]