how high up is our moon?

Answer 1

The average distance from the Earth to the Moon is about 384,400 kilometers (238,855 miles). This distance can vary slightly due to the Moon's elliptical orbit around Earth, ranging from about 363,300 kilometers (225,623 miles) at its closest (perigee) to about 405,500 kilometers (251,966 miles) at its farthest (apogee).

Answer 2

assume a moon with its own GBE and the fact its 238,855 miles in the air, how much energy would it take for character A to consistently move it out of orbit for 51 minutes?

Answer 3

To determine how much energy it would take for Character A to consistently move the Moon out of its natural orbit for 51 minutes, several factors need to be considered. These include the gravitational binding energy (GBE) of the Moon, its mass, the gravitational forces acting on it, and orbital mechanics.

Let’s break down the steps to calculate the required energy:

### 1. Gravitational Binding Energy (GBE)
The gravitational binding energy is the energy required to disperse an astronomical object into space. For the Moon, it can be approximated using the formula:

$$\text{GBE}=\frac{3G{M}^2}{5R}$$

Where:
- $G$ is the gravitational constant ($6.67430\times {10}^{-11}{\text{m}}^3{\text{kg}}^{-1}{\text{s}}^{-2}$),
- $M$ is the mass of the Moon ($7.347\times {10}^{22}\text{kg}$),
- $R$ is the radius of the Moon ($1.737\times {10}^6\text{m}$).

Plugging in these values:

$$\text{GBE}=\frac{3\times (6.67430\times {10}^{-11})\times (7.347\times {10}^{22}{)}^2}{5\times 1.737\times {10}^6}$$

$$\text{GBE}\approx 1.24\times {10}^{29}\text{J}$$

### 2. Energy to Change Orbit
Moving the Moon out of its natural orbit means overcoming its orbital velocity around Earth. The energy to completely change an orbit can be calculated using principles of orbital mechanics. Here, we assume Character A wants to move the Moon intermittently for 51 minutes.

We can also estimate the energy using the escape velocity concept:

$$E=\frac{1}{2}m{v}^2$$

Where:
- $m$ is the mass of the Moon,
- $v$ is the orbital velocity of the Moon around Earth ($\approx 1.022\text{km/s}$).

The kinetic energy for the Moon’s current orbit:

$$E\approx \frac{1}{2}\times 7.347\times {10}^{22}\times (1.022\times {10}^3{)}^2$$

$$E\approx \frac{1}{2}\times 7.347\times {10}^{22}\times 1.044\times {10}^6$$

$$E\approx 3.83\times {10}^{28}\text{J}$$

### 3. Work Done Over Time
If the energy needs to be maintained consistently over 51 minutes (3060 seconds), we can approximate the power (energy per unit time):

$$P=\frac{\text{Energy}}{\text{Time}}$$

To sustain the Moon out of its orbit for a precise period, continuous power is needed:

$$P\approx \frac{3.83\times {10}^{28}}{3060}$$

$$P\approx 1.25\times {10}^{25}\text{W}$$

This is an extremely rough estimate and doesn't cover many complex factors of Newtonian and Relativistic physics, like orbital stability and exact thrust requirements. The precise answer involves complex integrations and simulations beyond simple formulas.

**In summary**, Character A must provide a consistent energy output in the order of $1.25\times {10}^{25}$ watts over 51 minutes to keep the Moon dislocated from its natural orbit. This rough estimate highlights the incredibly high energy involved in such a task, reflecting the power and exertion required by Character A.

Answer 4

is that planet level or large planet level of energy output?

Answer 5

To determine whether the energy required to consistently move the Moon out of its orbit for 51 minutes falls under "planet level" or "large planet level" energy output, let’s compare it with some established benchmarks for these categories.

### Definitions and Benchmarks
- **Planet Level Energy:** Typically refers to the amount of energy required to destroy an Earth-sized planet or similar-sized celestial body, which is around ${10}^{32}$ joules. This is roughly the gravitational binding energy of Earth.
- **Large Planet Level Energy:** Refers to energies on the order of the gravitational binding energies of larger planets like Jupiter, which are substantially higher. For Jupiter, this energy is approximately $2\times {10}^{36}$ joules.

### Energy Required to Move the Moon
From the estimated power requirement:
$$P\approx 1.25\times {10}^{25}\text{W}$$
And the total energy output over 51 minutes (3060 seconds):
$$E=P\times t\approx 1.25\times {10}^{25}\text{W}\times 3060\text{s}$$
$$E\approx 3.83\times {10}^{28}\text{J}$$

### Comparison
- **Planet Level Energy:** $\approx {10}^{32}\text{J}$
- **Large Planet Level Energy:** $\approx {10}^{36}\text{J}$
- **Energy to Move the Moon for 51 minutes:** $\approx 3.83\times {10}^{28}\text{J}$

### Conclusion
The energy required to move the Moon for 51 minutes ($\approx 3.83\times {10}^{28}\text{J}$) is substantial, but it is still several orders of magnitude lower than the energy required for planetary destruction ($\approx {10}^{32}\text{J}$) or large planet-level energies ($\approx {10}^{36}\text{J}$).

Thus, the energy output required for Character A to move the Moon for 51 minutes would be categorized well below both "planet level" and "large planet level" energy outputs. It lies in a lower category, somewhere significantly less than planet-level by these common benchmarks.