Question

Your starship, the Aimless Wanderer, lands on the mysterious planet Mongo. As chief scientist-engineer, you make the following measurements: A 2.50 kg stone thrown upward from the ground at 13.0 m/s returns to the ground in 4.50 s; the circumference of Mongo at the equator is 2.00×10^5km; and there is no appreciable atmosphere on Mongo. The starship commander, Captain Confusion, asks for the following information:

a. What is the mass of Mongo?
b. If the Aimless Wanderer goes into a circular orbit 30,000 km above the surface of Mongo, how many hours will it take the ship to complete one orbit?

Answer 1

Given Information:  

Initial speed = v₁ = 13 m/s

time = t = 4.50 sec

Circumference of Mongo = C = 2.0×10⁵ km = 2.0×10⁸ m

Altitude = h = 30,000 km = 3×10⁷ m

Required Information:  

a) mass of Mongo = M = ?

b) time in hours = t = ?

Answer:  

a) mass of Mongo = M =  8.778×10²⁵ kg

b) time in hours = t = 11.08 h

Explanation:  

We know from the equations of kinematics,

v₂ = v₁t - ½gt²

0 = 13*4.50 - ½g(4.50)²

58.5 = 10.125g

g = 58.5/10.125

g = 5.78 m/s²

Newton's law of gravitation is given by

M = gC²/4π²G

Where C is the circumference of the planet Mongo, G is the gravitational constant, g is the acceleration of planet of Mongo and M is the mass of planet Mongo.

M = 5.78*(2.0×10⁸)²/(4π²*6.672×10⁻¹¹)

M = 8.778×10²⁵ kg

Therefore, the mass of planet Mongo is 8.778×10²⁵ kg

b) From the Kepler's third law,

T = 2π*(R + h)^3/2/(G*M)^1/2

Where R = C/2π

T = 2π*(C/2π + h)^3/2/(G*M)^1/2

T = 2π*((2.0×10⁸/2π) + 3×10⁷)^3/2/(6.672×10⁻¹¹*8.778×10²⁵)^1/2

T = 39917.5 sec

Convert to hours

T = 39917.5/60*60

T = 11.08 hours

Therefore, it will take 11.08 hours for the ship to complete one orbit.

Answer 2

(a) The mass of planet Mongo is  $8.778 \times 10^{25} \;\rm kg$.

(b) The time taken by the ship to complete on orbit is about 11.08 hours.

Given data:

The mass of stone is, m =2.50 kg.

The speed of stone is, u = 13.0 m/s.

The time taken to return to ground is, t = 4.50 s.

The circumference of Mongo at equator is, $C=2.00 \times 10^{5} \;\rm km$.

(a) Use Newton's law of gravitation to obtain the mass of gravitation as,

$M=\dfrac{g \times C^{2}}{4 \pi^{2}G}$ ........................................................(1)

Here,

g is the gravitational acceleration.

G is the universal gravitational constant.

The expression for the gravitational acceleration is,

$h=ut+\dfrac{1}{2}(-g)t^{2}\\\\0=13 \times 4.50+\dfrac{1}{2}(-g) \times (4.50)^{2}\\\\g = 5.78 \;\rm m/s^{2}$

Substitute the value in equation (1) as,

$M=\dfrac{5.78 \times (2 \times 10^{8})^{2}}{4 \pi^{2} \times 6.67 \times 10^{-11}}\\\\M = 8.778 \times 10^{25} \;\rm kg$

Thus, we can conclude that the mass of planet Mongo is  $8.778 \times 10^{25} \;\rm kg$.

(b)

Now apply the Kepler's law to obtain the time period for completing the one orbit as with the 30000 km height as,

$T=\dfrac{(R+h)^{3/2}}{GM^{1/2}}\\\\T=\dfrac{(C/2 \pi+h)^{3/2}}{GM^{1/2}}\\\\T=\dfrac{((2 \times 10^{8})/2 \pi+(3 \times 10^{7}))^{3/2}}{6.67 \times 10^{-11} \times (8.778 \times 10^{25})^{1/2}}\\\\T = 39917.5 \;\rm s =11.08 \;\rm h$

Thus, the time taken by the ship to complete one orbit is about 11.08 hours.

Learn more about the Kepler's Law here:

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