Find the work done in the propulsion problems.
4) Neglecting air resistance and the weight of the propellant, determine the work done in
propelling a five-ton satellite to a height of
a) 100 miles above Earth
b) 300 miles above Earth
Assume that the radius of Earth is 4000 miles.

To solve this propulsion problem, we will determine the work done to propel a satellite to given heights above Earth, neglecting air resistance and the weight of the propellant. We will use gravitational potential energy (GPE) to find the work required to move the satellite from the Earth's surface to a specific height. 

### Key Assumptions:
- Neglect air resistance and the weight of the propellant.
- The mass of the satellite is 5 tons (5 tons = 10,000 pounds = 10,000 lb).
- Radius of Earth \(R_E = 4000\) miles (converted to feet as needed).
- Use gravitational potential energy formula.

#### Work Done in Lifting a Satellite

The work done in lifting an object against gravity is the change in gravitational potential energy between two positions, given by:

\[
W = \Delta U = U_f - U_i
\]

where \(U_f\) is the final gravitational potential energy at height \(h\), and \(U_i\) is the initial gravitational potential energy on the surface of the Earth.

Gravitational potential energy \(U\) at a distance \(r\) from the center of the Earth is given by:

\[
U = -\frac{G M m}{r}
\]

Where:
- \(G = 6.674 \times 10^{-11} \, \text{N} \cdot \text{m}^2 / \text{kg}^2\) is the gravitational constant,
- \(M = 5.972 \times 10^{24} \, \text{kg}\) is the mass of the Earth,
- \(m\) is the mass of the satellite,
- \(r\) is the distance from the center of the Earth.

### Step-by-Step Calculation

For both cases (a) and (b), the work done will be the difference in potential energy between the final height and the initial height (which is the Earth's surface).

#### Part (a) Height = 100 miles above Earth

- Initial distance from Earth's center: \(r_i = R_E = 4000\) miles.
- Final distance from Earth's center: \(r_f = R_E + 100 = 4100\) miles.

#### Part (b) Height = 300 miles above Earth

- Initial distance from Earth's center: \(r_i = R_E = 4000\) miles.
- Final distance from Earth's center: \(r_f = R_E + 300 = 4300\) miles.

Let's calculate the work done in each case.

---

### Calculation for Part (a): \(h = 100\) miles
Converting miles to meters (as G, M, and \(r\) need to be in consistent SI units):

- \(1 \text{ mile} = 1609.34 \text{ meters}\)
- Radius of Earth \(R_E = 4000 \times 1609.34 = 6.374 \times 10^6 \text{ meters}\)
- Final distance \(r_f = 4100 \times 1609.34 = 6.599 \times 10^6 \text{ meters}\)

We'll now calculate the work done for both cases using these values.The work done to propel the five-ton satellite to the specified heights above Earth is as follows:

1. **Part (a): Height of 100 miles**
   - The work done is approximately \(6.85 \times 10^9\) joules (or 6.85 billion joules).

2. **Part (b): Height of 300 miles**
   - The work done is approximately \(1.96 \times 10^{10}\) joules (or 19.6 billion joules).

These values represent the energy required to lift the satellite from the Earth's surface to the respective heights.

Would you like further details or clarifications?

### Related Questions:
1. How would the work change if air resistance was included in the calculation?
2. What is the difference in energy if we were to double the mass of the satellite?
3. How does the work required change if the satellite is sent to geostationary orbit (22,236 miles)?
4. Can we calculate the velocity needed for the satellite to achieve orbit at these heights?
5. What is the effect of the Earth's rotation on the energy required?

### Tip:
In space propulsion problems, gravitational potential energy changes dominate for lower orbits, but kinetic energy becomes more significant for higher orbits and escapes velocity scenarios.

Neglecting air resistance and the weight of the propellant, determine the work done in propelling a five-ton satellite to a height of a) 100 miles above Earth b) 300 miles above Earth. Assume that the radius of Earth is 4000 miles.

Find the work done in lifting a five-ton satellite to 100 miles and 300 miles above Earth. This problem uses gravitational potential energy concepts, applying the formula U = -GMm/r. Suitable for advanced high school students in Grades 11-12.

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