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 upwardfrom the ground at 15.0 {\rm m/s} returns to the ground in 9.00 {\rm s}; the circumference of Mongo at the equator is1.00×105 {\rm km} ; and there is no appreciable atmosphere onMongo.part AThe starship commander, Captain Confusion,asks for the following information: what is the mass ofMongo?m ={\rm kg}part bIf the Aimless Wanderer goes into acircular orbit 2.00×104 {\rm km} above the surface of Mongo, how many hours will ittake the ship to complete one orbit?t ={\rm h}

Answer 1

a) M = 4,997 10²⁰ kg , b) T = 1.43 10³ s

Step-by-step explanation:

a) This exercise should be solved in several parts, let's start by calculating the acceleration of gravity of this planet from kinematics

v = v₀ - a t

As it indicates that there is no atmosphere, the friction force is zero and the initial and final velocity have the same module, but the opposite direction

a = (v₀ - v) / t

a = (15 - (-15)) /9.00 = 30/9

a = 3.33 m / s²

Now we use Newton's second law where force is the force of universal attraction

F = m a

G m M / r² = m a

M = a r² / G

Let's calculate

M = 3.33 (1.00 10⁵)² / 6.67 10⁻¹¹

M = 4,997 10²⁰ kg

b) The period of the ship's orbit

In this case we have a centripetal acceleration

The radius of the orbit is the radius of the plant plus the height of the ship from the surface

R =
`R_(m)` + h

R = 1 10⁵ + 2.00 10⁴

R = 12 10⁴ m

F = m a

G m M / R² = m a

Centripetal acceleration is

a = v² / R

The orbit is circular therefore the velocity module is constant, so we can use the equation of uniform motion, where the distance is the length of the orbit, for a circle

d = 2π R

v = d / t

v = 2π R / T

Let's replace

G m M / R² = m (2π R / T)² / R

G M = R³ 4π² / T²

T² = 4π² R³ / G M

T² = (4π² (12 10⁴)³ / (6.67 10⁻¹¹ 4,997 10²⁰)

T² = 6.82 10¹⁶ / 3.33 10¹⁰

T = √ (2,048 10⁶)

T = 1.43 10³ s