A 6000 - kg rocket is set for firing . If the exhaust speed is `1000 ms^(-1)` , how much gas must be ejected per second to supply the thrust needed , (a) to overcome the weight of the rocket , (b) to give the rocket an initial acceleration of `19.6 ms^(-2)`?

Options

A. 58.8 kg , 176.4
B. 60 kg ,176.4 kg
C. 75 kg , 180 kg
D. 90 kg , 120.3 kg

Answer

To solve the problem, we need to calculate the amount of gas that must be ejected per second to supply the required thrust for two scenarios: (a) to overcome the weight of the rocket and (b) to give the rocket an initial acceleration of \(19.6 \, \text{m/s}^2\).

### Given:
- Mass of the rocket, \( m_r = 6000 \, \text{kg} \)
- Exhaust speed, \( v_g = 1000 \, \text{m/s} \)
- Acceleration due to gravity, \( g = 9.8 \, \text{m/s}^2 \)

### (a) To overcome the weight of the rocket

1. **Calculate the weight of the rocket**:
\[ F_{\text{weight}} = m_r \cdot g = 6000 \, \text{kg} \cdot 9.8 \, \text{m/s}^2 = 58800 \, \text{N} \]

2. **Use the thrust equation**:
The thrust \( F_t \) produced by the rocket is given by the rate of change of momentum:
\[ F_t = \frac{dm}{dt} \cdot v_g \]
where \( \frac{dm}{dt} \) is the mass flow rate of the gas.

3. **Set thrust equal to weight**:
To overcome the weight, set the thrust equal to the weight:
\[ F_t = F_{\text{weight}} \implies \frac{dm}{dt} \cdot v_g = 58800 \, \text{N} \]

4. **Solve for \( \frac{dm}{dt} \)**:
\[ \frac{dm}{dt} = \frac{58800 \, \text{N}}{1000 \, \text{m/s}} = 58.8 \, \text{kg/s} \]

### (b) To give the rocket an initial acceleration of \(19.6 \, \text{m/s}^2\)

1. **Calculate the total force required**:
The total force required to give the rocket an acceleration \( a \) is given by:
\[ F_{\text{total}} = m_r \cdot g + m_r \cdot a \]
where \( a = 19.6 \, \text{m/s}^2 \).

2. **Substituting values**:
\[ F_{\text{total}} = 6000 \, \text{kg} \cdot 9.8 \, \text{m/s}^2 + 6000 \, \text{kg} \cdot 19.6 \, \text{m/s}^2 \]
\[ F_{\text{total}} = 58800 \, \text{N} + 117600 \, \text{N} = 176400 \, \text{N} \]

3. **Use the thrust equation**:
Again, using the thrust equation:
\[ F_t = \frac{dm}{dt} \cdot v_g \]

4. **Set thrust equal to total force**:
\[ \frac{dm}{dt} \cdot v_g = 176400 \, \text{N} \]

5. **Solve for \( \frac{dm}{dt} \)**:
\[ \frac{dm}{dt} = \frac{176400 \, \text{N}}{1000 \, \text{m/s}} = 176.4 \, \text{kg/s} \]

### Final Answers:
- (a) The mass flow rate to overcome the weight of the rocket is \( 58.8 \, \text{kg/s} \).
- (b) The mass flow rate to give the rocket an initial acceleration of \( 19.6 \, \text{m/s}^2 \) is \( 176.4 \, \text{kg/s} \).