Question

Determine the minimum amount of energy that a projectile of mass \(100.0 \mathrm{~kg}\) must gain to reach a circular orbit \(10.00 \mathrm{~km}\) above the Earth's surface if launched from (a) the North Pole or (b) the Equator (keep answers to four significant figures). Do not be concerned about the direction of the launch or of the final orbit. Is there an advantage or disadvantage to launching from the Equator? If so, how significant is the difference? Do not neglect the rotation of the Earth when calculating the initial energies. Use \(5.974 \cdot 10^{24} \mathrm{~kg}\) for the mass of the Earth and \(6357 \mathrm{~km}\) as the radius of the Earth.

Short answer

Question: Determine the minimum amount of energy needed for a projectile of mass 100 kg to reach a circular orbit 10.0 km above the Earth's surface, taking into account the projectile is launched from the North Pole and the Equator. Discuss the advantage or disadvantage of launching from the Equator.

Step-by-step solution

Calculate the initial energies

First, let's convert the distances to meters: R = 6357 km * 1000 = 6,357,000 m and h = 10,000 m. For the North Pole, the initial velocity \(v_{NP}\) of the projectile is assumed to be zero since there is no rotation in this case. Thus, the initial kinetic energy \(K_{NP}\) is also zero. For the Equator, the initial velocity of the projectile is equal to Earth's rotational velocity at the surface \(v_{EQ}\). We need to find the angular velocity \(\omega\) and calculate the linear velocity of the projectile as \(v_{EQ} = \omega R\). Earth makes a full rotation in 24 hours, which means \(\omega = \frac{2 \pi}{24 \cdot 60 \cdot 60} s^{-1}\). Now, we can calculate the initial kinetic energy \(K_{EQ}\) as \(K_{EQ} = \frac{1}{2} m v_{EQ}^2\).

Find gravitational potential energies

We need to find the gravitational potential energy at the Earth's surface and at the final orbit. For the Earth's surface, it is \(U_{0} = -\frac{GMm}{R}\). For the final orbit, it is \(U_{f} = -\frac{GMm}{R + h}\).

Determine the energy required

The energy required to reach a circular orbit from the North Pole is \(\Delta E_{NP} = (U_{f} + K_{f}) - (U_{0} + K_{NP})\), where \(K_{f} = \frac{1}{2} m v_{f}^2\) and \(v_{f} = \sqrt{\frac{GM}{R + h}}\). Similarly, the energy required to reach a circular orbit from the Equator is \(\Delta E_{EQ} = (U_{f} + K_{f}) - (U_{0} + K_{EQ})\).

Compare the energies and discuss

Now, let's compare the minimum energy required to reach a circular orbit when launched from the North Pole and the Equator: 1. Calculate the difference between required energies: \(\Delta E = \Delta E_{EQ} - \Delta E_{NP}\). 2. Discuss the advantage or disadvantage of launching from the Equator and the significance of the difference. After this step-by-step solution, you should be able to determine the minimum energy required for a projectile to reach a circular orbit above Earth, both from the North Pole and the Equator, and analyze the advantages or disadvantages of launching from the Equator.