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)`?

Answer

To solve the problem step by step, we will calculate the amount of gas that must be ejected per second to provide the necessary thrust for both 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 Data:
- Mass of the rocket, \(M_r = 6000 \, \text{kg}\)
- Exhaust speed, \(V_g = 1000 \, \text{m/s}\)
- Acceleration due to gravity, \(g = 9.8 \, \text{m/s}^2\)

### Part (a): To Overcome the Weight of the Rocket

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

2. **Set up the thrust equation:**
The thrust force \(F_t\) must equal the weight of the rocket to just lift it off the ground:
\[ F_t = M_r \cdot g \]
The thrust force can also be expressed in terms of the mass flow rate of the gas \( \frac{dM}{dt} \) and the exhaust speed:
\[ F_t = V_g \cdot \frac{dM}{dt} \]

3. **Equate the two expressions for thrust:**
\[ V_g \cdot \frac{dM}{dt} = M_r \cdot g \]

4. **Solve for the mass flow rate \( \frac{dM}{dt} \):**
\[ \frac{dM}{dt} = \frac{M_r \cdot g}{V_g} = \frac{6000 \, \text{kg} \cdot 9.8 \, \text{m/s}^2}{1000 \, \text{m/s}} = \frac{58800}{1000} = 58.8 \, \text{kg/s} \]

### Part (b): To Give the Rocket an Initial Acceleration of \(19.6 \, \text{m/s}^2\)

1. **Calculate the total thrust needed:**
The total thrust must overcome both the weight of the rocket and provide the additional force for acceleration:
\[ F_t = M_r \cdot (g + a) = 6000 \, \text{kg} \cdot (9.8 \, \text{m/s}^2 + 19.6 \, \text{m/s}^2) = 6000 \, \text{kg} \cdot 29.4 \, \text{m/s}^2 = 176400 \, \text{N} \]

2. **Set up the thrust equation again:**
\[ F_t = V_g \cdot \frac{dM}{dt} \]

3. **Equate the two expressions for thrust:**
\[ V_g \cdot \frac{dM}{dt} = M_r \cdot (g + a) \]

4. **Solve for the mass flow rate \( \frac{dM}{dt} \):**
\[ \frac{dM}{dt} = \frac{M_r \cdot (g + a)}{V_g} = \frac{6000 \, \text{kg} \cdot (9.8 \, \text{m/s}^2 + 19.6 \, \text{m/s}^2)}{1000 \, \text{m/s}} = \frac{6000 \cdot 29.4}{1000} = 176.4 \, \text{kg/s} \]

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