Answer :
Answer:
R = 7.915 10ā¶ m, M = 1.04 10Ā³āµ kg
Explanation:
Let's start by finding the acceleration of the planet's gravity, let's use the kinematic relations
v = vā - g t
the velocity of the body when it falls is the same for equal height, but it is positive when it rises and negative when it falls
v = -vā
-vā = vā - g t
g = 2vā / t
g = 2 15 / 2.7
g = 11.11 m / sĀ²
I now write the law of universal gravitation and Newton's second law
F = m a
G m M / RĀ² = m a
a = g
g = G M / RĀ²
Now let's work with the cruiser in orbit
F = ma
acceleration is centripetal
a = vĀ² / r
G m M / rĀ² = m vĀ² / r (1)
the distance from the center of the planet is
r = R + h
r = R + R = 2R
we substitute in 1
G M / 4RĀ² = vĀ² / 2R
G M / 2R = vĀ²
The modulus of the velocity in a circular orbit is
v = d / T
the distance is that of the circle
d = 2Ļ r
v = 2Ļ 2R / T
v = 4Ļ R / T
G M / 2R = 16piĀ² RĀ² / TĀ²
TĀ² = 32 piĀ² RĀ³ / GM
let's write the equations
g = G M / RĀ² (2)
TĀ² = 32 piĀ² RĀ³ / GM
we have two equations and two unknowns, so it can be solved
let's clear the most on the planet and equalize
g RĀ² / G = 32 piĀ² RĀ³ / GTĀ²
g TĀ² = 32 piĀ² R
R = g TĀ² / 32 piĀ²
let's reduce the period to SI units
T = 250 min (60 s / 1 min) = 1.5 104 s
let's calculate
R = 11.11 (1.5 10ā“) Ā² / 32 ĻĀ²
R = 7.915 10ā¶ m
from equation 2 we can find the mass of the planet
M = g RĀ² / G
M = 11.11 (7.915 10ā¶) Ā² / 6.67 10ā»Ā¹Ā¹
M = 1.04 10Ā³āµ kg