Obtain the rotational kinetic energy of the earth due its daily rotation on its axis. Assume the Earth to be a uniform sphere; take m = 5:98 x 10^(24) kg; r = 6:37 x10^6 m
1 point
9.8 x 10^(29)
2.56 x 10^(29)
8.85 x 10^(25)
1.00 x 10^(29)
KE=1/2 I w^2 where w is 2PI/day(change to seconds) and I=2mr^2/5
so change a day to seconds, and calculate.
KE=.5*.4*5.98e24*(6.37e6)^2 * (2*PI/8.64e4)^2
2.56*10^29J
Correct! The rotational kinetic energy of the Earth due to its daily rotation on its axis is approximately 2.56 x 10^29 J.
Well, what do you call a rotating Earth that's full of energy? A real spin doctor! Now, to calculate the rotational kinetic energy of the Earth, we can use the formula:
Rotational kinetic energy (KE) = (1/2) * I * ω^2
Where 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 * r^2, where m is the mass of the Earth and r is its radius.
Let's plug in the values: m = 5.98 x 10^24 kg and r = 6.37 x 10^6 m:
I = (2/5) * (5.98 x 10^24 kg) * (6.37 x 10^6 m)^2
Now, the angular velocity ω is given by ω = 2π / T, where T is the period of rotation. The Earth completes one rotation in approximately 24 hours, so T = 24 hours = 24 x 3600 seconds.
ω = 2π / (24 x 3600 s)
Now we can calculate ω and plug it into the formula above to find the rotational kinetic energy (KE) of the Earth. Let me do the math for you... *clownishly presses buttons*.
After crunching the numbers, the correct answer is 2.56 x 10^(29) joules. So, the second option is your best bet! Keep rocking and rolling with that energy calculation!
To obtain the rotational kinetic energy of the Earth due to its daily rotation on its axis, we can use the formula for rotational kinetic energy:
KE = (1/2) I ω²
where KE is the rotational kinetic energy, I is the moment of inertia, and ω is the angular velocity.
For a uniform sphere like the Earth, the moment of inertia (I) can be calculated using the equation:
I = (2/5) m r²
where m is the mass of the sphere (in this case, 5.98 x 10^24 kg) and r is its radius (in this case, 6.37 x 10^6 m).
The angular velocity (ω) can be calculated using:
ω = 2π / T
where T is the period of rotation (which is 24 hours for the Earth).
Now let's plug in the values to calculate the rotational kinetic energy:
I = (2/5) (5.98 x 10^24 kg) (6.37 x 10^6 m)² = 9.8251 x 10^37 kg m²
ω = 2π / (24 hours x 3600 seconds/hour) = 7.2921 x 10^(-5) rad/s
KE = (1/2) (9.8251 x 10^37 kg m²) (7.2921 x 10^(-5) rad/s)² = 2.5468 x 10^29 J
Therefore, the answer is 2.5468 x 10^29 J, so the correct option is 2.56 x 10^29.