You are an astronaut (m = 80 kg) and travel to a planet that is the same mass and size as Earth, but it has a rotational period of only 11 h. What is your apparent weight at the equator of this planet?
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To determine the apparent weight of the astronaut at the equator of this planet, we need to consider the rotational period of the planet and the gravitational force acting on the astronaut.
The rotational period of the planet is given as 11 hours. This means that it takes the planet 11 hours to complete one full rotation on its axis.
The formula to calculate the apparent weight of an object on a rotating planet is:
Apparent Weight = Actual Weight - Centrifugal Force
First, let's calculate the gravitational force acting on the astronaut. The formula for gravitational force is given by:
Gravitational Force = (Mass of the Astronaut) x (Acceleration due to Gravity)
On Earth, the acceleration due to gravity is approximately 9.8 m/s². Given that the mass of the astronaut is 80 kg, the gravitational force on Earth is:
Gravitational Force on Earth = (80 kg) x (9.8 m/s²) = 784 N
Next, let's calculate the Centrifugal Force. The Centrifugal Force is an apparent force that acts outward on a rotating object. It is given by the formula:
Centrifugal Force = (Mass of the Astronaut) x (Rotational Velocity)^2 / (Radius of the Planet)
The rotational velocity is the angular distance traveled per unit of time, given by:
Rotational Velocity = 2π / (Rotational Period)
The radius of the planet is not given, but we can assume it is the same as Earth, approximately 6,371 km (or 6,371,000 m).
Substituting these values into the Centrifugal Force formula, we get:
Rotational Velocity = 2π / (11 hours) = 2π / (11 × 60 × 60 seconds) = 1.45 x 10^(-4) rad/s
Centrifugal Force = (80 kg) x (1.45 x 10^(-4) rad/s)^2 / (6,371,000 m) = 0.078 N
Now, we can calculate the apparent weight:
Apparent Weight = Actual Weight - Centrifugal Force = 784 N - 0.078 N = 783.922 N
Therefore, the apparent weight of the astronaut at the equator of this planet would be approximately 783.922 Newtons.