Professor Frank Nabarro insists that all senior physics majors take his notorious physics aptitude test. The test is so tough th
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Answer:
a) M = 4,997 10²⁰ kg , b) T = 1.43 10³ s
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 = + 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