Question

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?

Answer 1

Answer:

Part a)

$M = 6.08 \times 10^{19} kg$

Part b)

$T = 4510 hours$

Explanation:

As we know that stone is thrown upwards with speed

$v_i = 12 m/s$

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

so the displacement of the stone is zero

$\Delta y = 0 = v t + \frac{1}{2}at^2$

$0 = 12 t - \frac{1}{2}g t^2$

$g = \frac{2(12)}{t}$

$g = 4 m/s^2$

Part a)

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

$L = 2 \times 100 km$

$2\pi R = 2\times 10^5 m$

$R = 3.2 \times 10^4 m$

Now we have formula of acceleration due to gravity as

$g = \frac{GM}{R^2}$

$4 = \frac{6.67 \times 10^{-11} M}{(3.2 \times 10^4)^2}$

$M = 6.08 \times 10^{19} kg$

Part b)

Time to complete one revolution around the planet is given as

$T = 2\pi\sqrt{\frac{r^3}{GM}}$

here we know that

r = distance from center of the planet

$r = 3.2 \times 10^4 + 3\times 10^7 = 3.003 \times 10^7 m$

now we have

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

$T = 4510 hours$