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 from (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.

Short answer

Please calculate the minimum amount of energy a 100.0 kg projectile needs to reach a circular orbit 10.00 km above Earth's surface when launched from the North Pole and the Equator, and compare the advantages or disadvantages of launching from the Equator. The minimum amount of energy required to reach the orbit from the North Pole is approximately 1.9698 x 10^10 J, whereas from the Equator, it is approximately 2.0778 x 10^10 J. Launching from the Equator requires about 1.0799 x 10^9 J less energy than launching from the North Pole due to Earth's rotation, which is roughly 5.19% less. This difference is significant, as it means launching from the Equator takes advantage of Earth's rotation to save energy, making it more efficient and practical for space launches.

Step-by-step solution

Gravitational potential energy at desired heights

First, we need to find the gravitational potential energy at the desired height (10.00 km above) for both launch cases. The formula for potential energy is: \(U_{g} = -\frac{G * m * M}{r}\), where \(G\) is the gravitational constant (\(6.674 \times 10^{-11} \frac{m^3}{kg s^{2}}\)), \(m\) is the mass of the projectile (100.0 kg), \(M\) is the mass of the Earth (\(5.972 \times 10^{24} kg\)), and \(r\) is the distance from the center of the Earth to the projectile. For both cases, the desired height is 10.00 km or 10000 m above the Earth's surface, add it to the Earth's radius (\(6.371 \times 10^6 m\)): \(r = R_{Earth} + h = 6.371 \times 10^6 m + 10000 m = 6.381 \times 10^6 m.\) Now calculate \(U_g\): \(U_{g} = -\frac{G * m * M}{r} = -\frac{6.674 \times 10^{-11} \frac{m^3}{kg s^{2}} * 100.0 kg * 5.972 \times 10^{24} kg}{6.381 \times 10^6 m} \approx -9.8696 \times 10^9 \text{ J}\)

Calculate kinetic energy for the projectile to maintain a circular orbit

Now we need to find the kinetic energy required for the projectile to maintain a circular orbit at the desired height. The formula for the kinetic energy is: \(K = \frac{1}{2}m v^2\), where \(m\) is the mass of the projectile, and \(v\) is the velocity needed to maintain the orbit. To find the velocity, we use the equation for the required orbital speed: \(v = \sqrt{\frac{G * M}{r}}\), Now, find the velocity using the given height: \(v = \sqrt{\frac{G * M}{r}} = \sqrt{\frac{6.674 \times 10^{-11} \frac{m^3}{kg s^{2}} * 5.972 \times 10^{24} kg}{6.381 \times 10^6 m}} \approx 7.6926 \times 10^3 \frac{m}{s}\) Now find the kinetic energy: \(K = \frac{1}{2}m v^2 = \frac{1}{2} * 100.0 kg * (7.6926 \times 10^3 \frac{m}{s})^2 \approx 2.9568 \times 10^{10} \text{ J}\)

Calculate the total energy for both North Pole and Equator launch cases

For the North Pole, we just need to add the gravitational potential energy and kinetic energy: \(E_{NP} = U_{g} + K \approx -9.8696 \times 10^9 \text{ J} + 2.9568 \times 10^{10} \text{ J} \approx 1.9698 \times 10^{10} \text{ J}\) For the Equator, we need to account for the Earth’s rotation, which provides an additional kinetic energy of: \(\frac{1}{2}m(v_{Rotation})^2\) The rotational velocity at the equator is, \(v_{Rotation} = R_{Earth} * \omega = R_{Earth} * \frac{2\pi}{T} = 6.371 \times 10^6 m * \frac{2\pi}{24*60*60 s} \approx 465.18 \frac{m}{s}\) Now, find the additional kinetic energy due to Earth's rotation: \(K_{Rotation} = \frac{1}{2}m(v_{rotation})^2 = \frac{1}{2} * 100.0 kg * (465.18 \frac{m}{s})^2 \approx 1.0799 \times 10^9 \text{ J}\) Now, add \(K_{Rotation}\) to the total energy for the Equator: \(E_{EQ} = E_{NP} + K_{Rotation} \approx 1.9698 \times 10^{10} \text{ J} + 1.0799 \times 10^9 \text{ J} \approx 2.0778 \times 10^{10} \text{ J}\)

Compare the energies and discuss the significance of launching from the Equator

Now, let's compare the energies required for launches at both locations: \(E_{NP} \approx 1.9698 \times 10^{10} \text{ J}\) \(E_{EQ} \approx 2.0778 \times 10^{10} \text{ J}\) Launching from the Equator requires about \(1.0799 \times 10^9 \text{ J}\) less energy than launching from the North Pole due to Earth's rotation, which is roughly 5.19% less. This difference is significant, as it means launching from the Equator takes advantage of Earth's rotation to save energy, making it more efficient and practical for space launches.

Key concepts

Gravitational Potential Energy

Gravitational potential energy (GPE) is the energy an object possesses because of its position in a gravitational field. This concept is crucial in physics and is particularly significant when discussing satellite launches, as the GPE must be overcome to place an object in orbit.

The formula to calculate GPE is \(U_{g} = -\frac{G * m * M}{r}\), where \(G\) is the universal gravitational constant, \(m\) is the mass of the projectile, \(M\) is the mass of the Earth, and \(r\) is the distance from the center of the Earth to the projectile, just beyond Earth's atmosphere.

Why Negative Value?

The potential energy is considered negative because gravitational force is attractive; energy is released when the object falling towards Earth moves to a lower, more stable energy state. To launch a projectile into orbit, adding the required kinetic energy turns this negative value into a positive total energy state, which is necessary for maintaining an orbit.

For a projectile launched from Earth’s surface, it's essential to count in the additional height when it reaches its intended orbit. This calculation shows how much work needs to be done against Earth's gravity to elevate an object to that particular height.

Circular Orbit Velocity

Circular orbit velocity is the speed a satellite or any object must maintain to orbit a celestial body in a circular path. Ensuring that the object remains in orbit without spiraling into the planet or drifting into space is a balance of forces. The formula is \(v = \sqrt{\frac{G * M}{r}}\), where \(v\) is the orbital velocity, \(G\) is the gravitational constant, \(M\) the mass of the celestial body, and \(r\) the radius of the circular orbit.

Understanding the Orbit Speed

The speed calculation for a circular orbit velocity illustrates the 'just-right' velocity needed for an object to counteract the pull of gravity while avoiding being accelerated towards the Earth or flung out of orbit. It reflects a balance between gravitational and inertial forces where the centrifugal force due to the object's motion counters the gravitational pull from the Earth.

For students, envisioning the orbit speed as 'the speed you need to go so that you fall towards the Earth but the Earth curves away underneath you at the same rate' may help in grasping this delicate balance.

Earth's Rotation Effect on Space Launch

Earth's rotation can provide a significant advantage for space launches when the launch is from the equator. As the Earth spins, it imparts a rotational velocity to any object on its surface. For launches attaining orbit, this rotation can be used to boost the projectile and, in effect, reduce the energy required to reach orbit.

The rotational speed is calculated by \(v_{Rotation} = R_{Earth} * \omega = R_{Earth} * \frac{2\pi}{T}\), with \(\omega\) being the angular velocity, and \(T\) the period of Earth's rotation, which is a day. The effective kinetic energy gained from Earth's rotation is \(K_{Rotation} = \frac{1}{2}m(v_{Rotation})^2\).

Efficiency at the Equator

At the equator, the rotational speed of Earth is at its maximum, thus providing the greatest potential kinetic energy to assists launch. This additional speed means we require less fuel, which translates to lower costs and a smaller impact on the environment. This is a key reason why space agencies often prefer equatorial launch sites. Sites like the Guiana Space Centre are so coveted within the aerospace industry as they can launch more massive payloads or save significant fuel thanks in part to Earth's rotational boost.