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

Part a)

`M = 6.08 * 10^(19) kg`

Part b)

`T = 4510 hours`

Step-by-step 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 + (1)/(2)at^2`

`0 = 12 t - (1)/(2)g t^2`

`g = (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 * 100 km`

`2\pi R = 2* 10^5 m`

`R = 3.2 * 10^4 m`

Now we have formula of acceleration due to gravity as

`g = (GM)/(R^2)`

`4 = (6.67 * 10^(-11) M)/((3.2 * 10^4)^2)`

`M = 6.08 * 10^(19) kg`

Part b)

Time to complete one revolution around the planet is given as

`T = 2\pi\sqrt{(r^3)/(GM)}`

here we know that

r = distance from center of the planet

`r = 3.2 * 10^4 + 3* 10^7 = 3.003 * 10^7 m`

now we have

`T = 2\pi\sqrt{((3.003* 10^7)^3)/((6.67 * 10^(-11))(6.08* 10^(19)))}`

`T = 4510 hours`