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Answer:
13.6 cm
Explanation:
From Snell's law:
n₁ sin θ₁ = n₂ sin θ₂
In the air, n₁ = 1, and light from the horizon forms a 90° angle with the vertical, so sin θ₁ = sin 90° = 1.
Given n₂ = 4/3:
1 = 4/3 sin θ
sin θ = 3/4
If x is the radius of the circle, then sin θ is:
sin θ = x / √(x² + 12²)
sin θ = x / √(x² + 144)
Substituting:
3/4 = x / √(x² + 144)
9/16 = x² / (x² + 144)
9/16 x² + 81 = x²
81 = 7/16 x²
x ≈ 13.6
A.
The orbital speed of the clumps of matter around the black hole is equal to the ratio between the circumference of the orbit and the period of revolution:
where we have:
is the orbital speed
r is the orbital radius
is the orbital period
Solving for r, we find the distance of the clumps of matter from the centre of the black hole:
B.
The gravitational force between the black hole and the clumps of matter provides the centripetal force that keeps the matter in circular motion:
where
m is the mass of the clumps of matter
G is the gravitational constant
M is the mass of the black hole
Solving the formula for M, we find the mass of the black hole:
and considering the value of the solar mass
the mass of the black hole as a multiple of our sun's mass is
C.
The radius of the event horizon is equal to the Schwarzschild radius of the black hole, which is given by
where M is the mass of the black hole and c is the speed of light.
Substituting numbers into the formula, we find
Complete question:
In the movie The Martian, astronauts travel to Mars in a spaceship called Hermes. This ship has a ring module that rotates around the ship to create “artificial gravity” within the module. Astronauts standing inside the ring module on the outer rim feel like they are standing on the surface of the Earth. (The trailer for this movie shows Hermes at t=2:19 and demonstrates the “artificial gravity” concept between t= 2:19 and t=2:24.)
Analyzing a still frame from the trailer and using the height of the actress to set the scale, you determine that the distance from the center of the ship to the outer rim of the ring module is 11.60 m
What does the rotational speed of the ring module have to be so that an astronaut standing on the outer rim of the ring module feels like they are standing on the surface of the Earth?
Answer:
The rotational speed of the ring module have to be 0.92 rad/s
Explanation:
Given;
the distance from the center of the ship to the outer rim of the ring module r, = 11.60 m
When the astronaut standing on the outer rim of the ring module feels like they are standing on the surface of the Earth, then their centripetal acceleration will be equal to acceleration due to gravity of Earth.
Centripetal acceleration, a = g = 9.8 m/s²
Centripetal acceleration, a = v²/r
But v = ωr
a = g = ω²r
Therefore, the rotational speed of the ring module have to be 0.92 rad/s