A body is projected vertically upward from the surface of the earth with escape velocity. Calculate the time in which it will be at a height (measured from the surface of the arth) 8 time the radius of the earth (R). Acceleration due to gravity on the surface of the earth is g.

Answer

To solve the problem, we need to calculate the time taken for a body projected vertically upward from the surface of the Earth with escape velocity to reach a height of 8 times the radius of the Earth (8R). The acceleration due to gravity at the surface of the Earth is denoted as g.

### Step-by-Step Solution:

1. **Understanding Escape Velocity**:
The escape velocity (\(v_e\)) from the surface of the Earth is given by the formula:
\[ v_e = \sqrt{\frac{2GM}{R}} \]
where \(G\) is the gravitational constant, \(M\) is the mass of the Earth, and \(R\) is the radius of the Earth.

2. **Energy Conservation**:
When the body is projected with escape velocity, the total mechanical energy at the surface is zero:
\[ \frac{1}{2} mv_e^2 - \frac{GMm}{R} = 0 \]
This implies that the kinetic energy equals the gravitational potential energy at the surface.

3. **Velocity at Height \(x\)**:
At a distance \(x\) from the center of the Earth, the velocity \(V\) of the body can be expressed as:
\[ \frac{1}{2} mV^2 - \frac{GMm}{x} = 0 \]
Rearranging gives:
\[ V = \sqrt{\frac{2GM}{x}} \]

4. **Relating Velocity and Time**:
The velocity can also be expressed as:
\[ V = \frac{dx}{dt} \]
Therefore, we have:
\[ \sqrt{\frac{2GM}{x}} = \frac{dx}{dt} \]
Rearranging gives:
\[ dt = \frac{dx}{\sqrt{\frac{2GM}{x}}} = \sqrt{\frac{x}{2GM}} dx \]

5. **Integrating to Find Time**:
To find the time taken to reach a height of \(8R\) (which corresponds to a distance of \(9R\) from the center of the Earth), we need to integrate from \(R\) to \(9R\):
\[ t = \int_{R}^{9R} \sqrt{\frac{x}{2GM}} dx \]
The integral simplifies to:
\[ t = \sqrt{\frac{1}{2GM}} \int_{R}^{9R} \sqrt{x} dx \]

6. **Calculating the Integral**:
The integral \(\int \sqrt{x} dx = \frac{2}{3} x^{3/2}\). Evaluating from \(R\) to \(9R\):
\[ \int_{R}^{9R} \sqrt{x} dx = \left[ \frac{2}{3} x^{3/2} \right]_{R}^{9R} = \frac{2}{3} \left( (9R)^{3/2} - R^{3/2} \right) \]
Simplifying gives:
\[ = \frac{2}{3} \left( 27R^{3/2} - R^{3/2} \right) = \frac{2}{3} \cdot 26R^{3/2} = \frac{52}{3} R^{3/2} \]

7. **Final Expression for Time**:
Substituting back into the expression for \(t\):
\[ t = \sqrt{\frac{1}{2GM}} \cdot \frac{52}{3} R^{3/2} \]
We know that \(\frac{GM}{R^2} = g\), so:
\[ t = \frac{52}{3} \sqrt{\frac{R}{2g}} \]

### Final Answer:
The time taken for the body to reach a height of \(8R\) from the surface of the Earth is:
\[ t = \frac{52}{3} \sqrt{\frac{R}{2g}} \]