Calculate the velocity with which a body must be thrown vertically upward from the surface of the earth so that it may reach a height of `10R`, where `R` is the radius of the earth and is equal to `6.4 xx 10^(6)m.` (Given: Mass of the earth `= 6 xx 10^(24) kg`, gravitational constant `G = 6.7 xx 10^(-11) N m^(2) kg^(-2)`)

Explanation

To solve the problem of calculating the velocity with which a body must be thrown vertically upward from the surface of the Earth to reach a height of \(10R\), we can use the principle of conservation of energy. Here’s a step-by-step solution:

### Step 1: Understand the Problem
We need to find the initial velocity \(v\) required for a body to reach a height of \(10R\) above the Earth's surface, where \(R\) is the radius of the Earth.

### Step 2: Use Conservation of Energy
The total mechanical energy at the surface of the Earth (initial state) must equal the total mechanical energy at the height \(10R\) (final state). The equation can be set up as follows:

\[ \frac{1}{2}mv^2 - \frac{GMm}{R} = -\frac{GMm}{R + 10R} \]

Where:
- \(m\) is the mass of the body,
- \(M\) is the mass of the Earth,
- \(G\) is the gravitational constant,
- \(R\) is the radius of the Earth.

### Step 3: Simplify the Equation
We can simplify the equation by canceling \(m\) from both sides (assuming \(m \neq 0\)):

\[ \frac{1}{2}v^2 - \frac{GM}{R} = -\frac{GM}{11R} \]

### Step 4: Rearrange the Equation
Rearranging the equation gives us:

\[ \frac{1}{2}v^2 = \frac{GM}{R} - \frac{GM}{11R} \]

### Step 5: Combine the Terms
Combining the terms on the right side:

\[ \frac{1}{2}v^2 = GM \left( \frac{1}{R} - \frac{1}{11R} \right) \]

\[ \frac{1}{2}v^2 = GM \left( \frac{11 - 1}{11R} \right) = \frac{10GM}{11R} \]

### Step 6: Solve for \(v^2\)
Multiplying both sides by 2:

\[ v^2 = \frac{20GM}{11R} \]

### Step 7: Calculate \(v\)
Taking the square root gives us:

\[ v = \sqrt{\frac{20GM}{11R}} \]

### Step 8: Substitute Known Values
Now, we will substitute the known values:
- \(G = 6.7 \times 10^{-11} \, \text{N m}^2/\text{kg}^2\)
- \(M = 6 \times 10^{24} \, \text{kg}\)
- \(R = 6.4 \times 10^6 \, \text{m}\)

Substituting these values into the equation:

\[ v = \sqrt{\frac{20 \times (6.7 \times 10^{-11}) \times (6 \times 10^{24})}{11 \times (6.4 \times 10^6)}} \]

### Step 9: Calculate the Numerical Value
Calculating the above expression step-by-step:

1. Calculate \(20 \times 6.7 \times 10^{-11} \times 6 \times 10^{24}\):
\[ = 8.04 \times 10^{14} \]

2. Calculate \(11 \times 6.4 \times 10^6\):
\[ = 7.04 \times 10^7 \]

3. Now divide:
\[ \frac{8.04 \times 10^{14}}{7.04 \times 10^7} \approx 1.14 \times 10^7 \]

4. Finally, take the square root:
\[ v \approx \sqrt{1.14 \times 10^7} \approx 1.07 \times 10^4 \, \text{m/s} \]

### Final Answer
The velocity with which the body must be thrown is approximately \(1.07 \times 10^4 \, \text{m/s}\).

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