By how much percent does the speed of a satellite orbiting in circular orbit be increased so that it will escape from the gravitational field of the earth ?

Options

A. `73.2%`
B. `41.4%`
C. `21.6%`
D. `66.6%`

Answer

To solve the problem of how much percent the speed of a satellite in a circular orbit must be increased to escape the gravitational field of the Earth, we can follow these steps:

### Step 1: Understand the formulas for orbital speed and escape velocity
The orbital speed \( v_o \) of a satellite in a circular orbit is given by the formula:
\[ v_o = \sqrt{\frac{GM}{R}} \]
where \( G \) is the gravitational constant, \( M \) is the mass of the Earth, and \( R \) is the radius of the orbit.

The escape velocity \( v_e \) from the gravitational field is given by:
\[ v_e = \sqrt{\frac{2GM}{R}} \]

### Step 2: Relate escape velocity to orbital speed
From the above formulas, we can see that:
\[ v_e = \sqrt{2} \cdot v_o \]
This means that the escape velocity is \( \sqrt{2} \) times the orbital speed.

### Step 3: Calculate the increase in speed
To find the increase in speed required, we can express the increase in speed as:
\[ \Delta v = v_e - v_o = \sqrt{2} \cdot v_o - v_o \]
Factoring out \( v_o \):
\[ \Delta v = v_o (\sqrt{2} - 1) \]

### Step 4: Calculate the percentage increase
The percentage increase in speed is given by:
\[ \text{Percentage Increase} = \left( \frac{\Delta v}{v_o} \right) \times 100\% \]
Substituting for \( \Delta v \):
\[ \text{Percentage Increase} = \left( \frac{v_o (\sqrt{2} - 1)}{v_o} \right) \times 100\% = (\sqrt{2} - 1) \times 100\% \]

### Step 5: Calculate the numerical value
Now we can calculate \( \sqrt{2} \):
\[ \sqrt{2} \approx 1.414 \]
Thus:
\[ \text{Percentage Increase} = (1.414 - 1) \times 100\% = 0.414 \times 100\% = 41.4\% \]

### Conclusion
The speed of the satellite must be increased by approximately **41.4%** in order to escape the gravitational field of the Earth.

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