A rocket of mass 5000 kg is to be projected vertically upward. The gases are exhausted vertically downwards with velocity `1000ms^(-1)` with respect to the rocket. What is the minimum rate of burning the fuel so as to just lift the rocket upwards against gravitational attraction ?

Options

A. `49kgs^(-1)`
B. `147kgs^(-1)`
C. `98kgs^(-1)`
D. `196kgs^(-1)`

Explanation

To solve the problem of determining the minimum rate of burning fuel for a rocket to just lift off against gravitational attraction, we can follow these steps:

### Step 1: Identify the forces acting on the rocket
The forces acting on the rocket are:
- The gravitational force (weight) acting downwards, which is given by \( mg \), where \( m \) is the mass of the rocket and \( g \) is the acceleration due to gravity (approximately \( 9.8 \, \text{m/s}^2 \)).
- The thrust force acting upwards due to the exhaust gases being expelled downwards.

### Step 2: Set up the equilibrium condition for lift-off
For the rocket to just lift off, the upward thrust force must equal the downward gravitational force:
\[ F_{\text{thrust}} = mg \]

### Step 3: Calculate the thrust force
The thrust force can be expressed in terms of the mass flow rate of the exhaust gases and their velocity. If \( \frac{dm}{dt} \) is the rate of burning fuel (mass flow rate), and \( v_{\text{exhaust}} \) is the velocity of the exhaust gases with respect to the rocket, then the thrust force is given by:
\[ F_{\text{thrust}} = \frac{dm}{dt} \cdot v_{\text{exhaust}} \]

### Step 4: Substitute the known values
From the problem, we know:
- Mass of the rocket, \( m = 5000 \, \text{kg} \)
- Velocity of exhaust gases, \( v_{\text{exhaust}} = 1000 \, \text{m/s} \)
- Acceleration due to gravity, \( g = 9.8 \, \text{m/s}^2 \)

Substituting these into the equilibrium condition:
\[ \frac{dm}{dt} \cdot v_{\text{exhaust}} = mg \]
\[ \frac{dm}{dt} \cdot 1000 = 5000 \cdot 9.8 \]

### Step 5: Solve for the mass flow rate \( \frac{dm}{dt} \)
Rearranging the equation to solve for \( \frac{dm}{dt} \):
\[ \frac{dm}{dt} = \frac{5000 \cdot 9.8}{1000} \]
\[ \frac{dm}{dt} = \frac{49000}{1000} = 49 \, \text{kg/s} \]

### Conclusion
The minimum rate of burning the fuel so as to just lift the rocket upwards against gravitational attraction is \( 49 \, \text{kg/s} \).

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