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Sagot :
Answer:
R = 7.915 10⁶ m, M = 1.04 10³⁵ kg
Explanation:
Let's start by finding the acceleration of the planet's gravity, let's use the kinematic relations
v = v₀ - 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 = -v₀
-v₀ = v₀ - g t
g = 2v₀ / t
g = 2 15 / 2.7
g = 11.11 m / s²
I now write the law of universal gravitation and Newton's second law
F = m a
G m M / R² = m a
a = g
g = G M / R²
Now let's work with the cruiser in orbit
F = ma
acceleration is centripetal
a = v² / r
G m M / r² = m v² / r (1)
the distance from the center of the planet is
r = R + h
r = R + R = 2R
we substitute in 1
G M / 4R² = v² / 2R
G M / 2R = v²
The modulus of the velocity in a circular orbit is
v = d / T
the distance is that of the circle
d = 2π r
v = 2π 2R / T
v = 4π R / T
G M / 2R = 16pi² R² / T²
T² = 32 pi² R³ / GM
let's write the equations
g = G M / R² (2)
T² = 32 pi² R³ / GM
we have two equations and two unknowns, so it can be solved
let's clear the most on the planet and equalize
g R² / G = 32 pi² R³ / GT²
g T² = 32 pi² R
R = g T² / 32 pi²
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 (1.5 10⁴) ² / 32 π²
R = 7.915 10⁶ m
from equation 2 we can find the mass of the planet
M = g R² / G
M = 11.11 (7.915 10⁶) ² / 6.67 10⁻¹¹
M = 1.04 10³⁵ kg
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