A 6.80 x 10^4 kg space shuttle moving at 7.90 x 10^4 m/s reenters the earth's atmosphere 80.0 km above the earth's surface. What is the shuttle's total mechanical energy with respect to earth's surface?
I know the answer is E= 2.173 x 10^12J
But I don't know where this answer comes from. I tried using the formula E= K + Ug
Ug = mgh
What am I doing wrong?
The kinetic energy alone is
(1/2) m v^2 = 212 * 10^12
so there is a typo somewhere either in the problem statement or in the answer.
E= 2.173 × 10^12 J
E= 2.173 ×10^12 J
Please show the state to me
To find the shuttle's total mechanical energy with respect to the Earth's surface, we need to consider its kinetic energy (K) and potential energy (Ug).
The formula you mentioned, E = K + Ug, is correct. However, you need to be careful about how you calculate the potential energy (Ug) in this case.
The potential energy is given by the formula Ug = mgh, where:
- m is the mass of the object,
- g is the acceleration due to gravity, and
- h is the height above the reference point.
In this problem, the space shuttle is already 80.0 km above the Earth's surface, so the height (h) is given as 80.0 km. However, you need to convert it to meters before using it in the equation.
From the given values, we have:
- m = 6.80 x 10^4 kg (mass of the space shuttle)
- h = 80.0 km = 80.0 x 10^3 m (height above the Earth's surface)
Now, you can calculate the potential energy:
Ug = mgh = (6.80 x 10^4 kg) x (9.8 m/s^2) x (80.0 x 10^3 m) = ...
After calculating Ug, you can substitute the values of K and Ug into the energy equation (E = K + Ug) to find the total mechanical energy.
Remember to convert any units as necessary and perform the calculations accurately to obtain the correct final answer.