Calculate the kinetic energy that the earth has because of (a) its rotation about its own axis and (b) its motion around the sun. Assume that the earth is a uniform sphere and that its path around the sun is circular. For comparison, the total energy used in the United States in one year is about 1.1 x 1020 J.
(a) KE(rot) =I ω²/2
I=2MR²/5
ω=2π/24•3600 = ...(rad/ s)
Earth’s mass is M = 5.97•10²⁴kg,
Earth’s radius is R = 6.378•10⁶ m.
(b)
Linear velocity of the Earth is
v=ωr = 2π• 1.496•10¹¹ /365•24•3600 = …m/s
KE=Mv²/2 =5.97•10²⁴• (2π• 1.496•10¹¹ /365•24•3600)²/2 = …J
Thanks!! Clears up everything
To calculate the kinetic energy of the Earth due to its rotation about its own axis, we can use the formula:
Kinetic Energy = (1/2) * I * ω²
where I is the moment of inertia and ω is the angular velocity.
Given that the Earth is a uniform sphere, the moment of inertia (I) is given by:
I = (2/5) * M * R²
where M is the mass of the Earth and R is the radius of the Earth.
The angular velocity (ω) can be calculated by dividing the angular displacement (2π radians) by the time taken for one rotation (24 hours).
Let's calculate the values:
Mass of the Earth (M) = 5.972 x 10^24 kg
Radius of the Earth (R) = 6.371 x 10^6 m
Time taken for one rotation (t) = 24 hours = 24 x 60 x 60 seconds
Using the above values, we can calculate:
I = (2/5) * M * R²
= (2/5) * (5.972 x 10^24 kg) * (6.371 x 10^6 m)²
ω = 2π / t
= 2π / (24 x 60 x 60 seconds)
Finally, we can calculate the kinetic energy (KE) due to rotation:
KE = (1/2) * I * ω²
Now, let's move on to calculating the kinetic energy of the Earth due to its motion around the Sun.
The kinetic energy of the Earth due to its motion around the Sun is given by:
KE = (1/2) * M * v²
where M is the mass of the Earth and v is the orbital velocity.
The orbital velocity can be calculated using the formula:
v = 2πR / T
where R is the average distance from the Earth to the Sun, and T is the time period of one revolution around the Sun.
Given that the Earth's path around the Sun is approximately a circle, we can assume the radius of this circular path is the average distance between the Earth and the Sun.
Let's calculate the values:
Average distance between the Earth and the Sun (R) = 1.496 x 10^11 m
Time period of one revolution around the Sun (T) = 365 days = 365 x 24 x 60 x 60 seconds
Using the above values, we can calculate:
v = 2πR / T
Finally, we can calculate the kinetic energy (KE) due to motion around the Sun:
KE = (1/2) * M * v²
By calculating both the rotational kinetic energy and orbital kinetic energy, we can determine the total kinetic energy of the Earth.
To calculate the kinetic energy of the Earth due to its rotation about its own axis, we can use the formula for the kinetic energy of a rotating object:
KE = (1/2) I ω²
Where KE is the kinetic energy, I is the moment of inertia, and ω is the angular velocity.
For a uniform sphere, the moment of inertia is given by:
I = (2/5) m r²
Where m is the mass of the Earth, and r is its radius.
The angular velocity of the Earth's rotation can be calculated as:
ω = 2π / T
Where T is the period of the Earth's rotation, which is approximately 24 hours or 86,400 seconds.
Plugging in the values, we get:
I = (2/5) (5.97 x 10²⁴ kg) (6.37 x 10⁶ m)²
≈ 9.93 x 10³⁶ kg m²
ω = 2π / (24 x 60 x 60 s)
≈ 7.27 x 10⁻⁵ rad/s
Substituting the values into the formula for kinetic energy, we find:
KE = (1/2) (9.93 x 10³⁶ kg m²) (7.27 x 10⁻⁵ rad/s)²
≈ 2.14 x 10²⁹ J
So the kinetic energy of the Earth due to its rotation about its own axis is approximately 2.14 x 10²⁹ Joules.
Now let's calculate the kinetic energy of the Earth due to its motion around the sun. Since the path of the Earth around the sun is assumed to be circular, we can use the formula for the kinetic energy of a circular motion:
KE = (1/2) m v²
Where m is the mass of the Earth, and v is its orbital velocity.
The mass of the Earth is 5.97 x 10²⁴ kg.
The orbital velocity can be calculated as:
v = 2π r / T
Where r is the average distance between the Earth and the sun, and T is the period of the Earth's orbit, which is approximately 365.25 days or 3.15 x 10⁷ seconds.
The average distance between the Earth and the sun is approximately 1.496 x 10¹¹ meters.
Plugging in the values, we get:
v = (2π) (1.496 x 10¹¹ m) / (3.15 x 10⁷ s)
≈ 2.98 x 10⁴ m/s
Substituting the values into the formula for kinetic energy, we find:
KE = (1/2) (5.97 x 10²⁴ kg) (2.98 x 10⁴ m/s)²
≈ 8.91 x 10³³ J
So the kinetic energy of the Earth due to its motion around the sun is approximately 8.91 x 10³³ Joules.
Therefore, the total kinetic energy of the Earth is the sum of the kinetic energies due to its rotation and its motion around the sun:
Total KE = (2.14 x 10²⁹ J) + (8.91 x 10³³ J)
≈ 8.91 x 10³³ J
For comparison, the total energy used in the United States in one year is about 1.1 x 10²⁰ J, which is significantly lower than the kinetic energy of the Earth.