A particle is projected radially outwards from the surface of a planet with an initial speed of `4.0km//s`. Maximum height attained by the particle is `h=900km`. Find the radius of the planet taking `g=10m//s^(2)`.

Answer

To solve the problem, we will use the conservation of mechanical energy principle. The total mechanical energy (kinetic + potential) at the surface of the planet will be equal to the total mechanical energy at the maximum height.

### Step-by-Step Solution:

1. **Identify the Known Values:**
- Initial speed \( v_i = 4.0 \, \text{km/s} = 4000 \, \text{m/s} \)
- Maximum height \( h = 900 \, \text{km} = 900000 \, \text{m} \)
- Acceleration due to gravity \( g = 10 \, \text{m/s}^2 \)

2. **Write the Conservation of Energy Equation:**
At the surface of the planet:
- Initial kinetic energy (KE) = \( \frac{1}{2} m v_i^2 \)
- Initial potential energy (PE) = \( -\frac{GMm}{R} \)

At the maximum height:
- Final kinetic energy (KE) = 0 (since the particle stops momentarily)
- Final potential energy (PE) = \( -\frac{GMm}{R + h} \)

The conservation of energy gives us:
\[ \frac{1}{2} m v_i^2 - \frac{GMm}{R} = -\frac{GMm}{R + h} \]

3. **Simplify the Equation:**
Canceling \( m \) from both sides (assuming \( m \neq 0 \)):
\[ \frac{1}{2} v_i^2 - \frac{GM}{R} = -\frac{GM}{R + h} \]

Rearranging gives:
\[ \frac{1}{2} v_i^2 = \frac{GM}{R} - \frac{GM}{R + h} \]

4. **Combine the Terms:**
The right-hand side can be combined:
\[ \frac{GM}{R} - \frac{GM}{R + h} = GM \left( \frac{1}{R} - \frac{1}{R + h} \right) = GM \left( \frac{(R + h) - R}{R(R + h)} \right) = \frac{GMh}{R(R + h)} \]

Thus, we have:
\[ \frac{1}{2} v_i^2 = \frac{GMh}{R(R + h)} \]

5. **Substitute \( GM \) Using \( g \):**
From the formula \( g = \frac{GM}{R^2} \), we can express \( GM \) as:
\[ GM = gR^2 \]

Substituting this into our equation gives:
\[ \frac{1}{2} v_i^2 = \frac{gR^2h}{R(R + h)} = \frac{gRh}{R + h} \]

6. **Rearranging for \( R \):**
Rearranging gives:
\[ R + h = \frac{gRh}{\frac{1}{2} v_i^2} \]
\[ R + h = \frac{2gRh}{v_i^2} \]
\[ R(1 - \frac{2gh}{v_i^2}) = -h \]
\[ R = \frac{h}{\frac{2gh}{v_i^2} - 1} \]

7. **Substituting the Values:**
Substitute \( g = 10 \, \text{m/s}^2 \), \( h = 900000 \, \text{m} \), and \( v_i = 4000 \, \text{m/s} \):
\[ R = \frac{900000}{\frac{2 \times 10 \times 900000}{4000^2} - 1} \]

Calculate \( \frac{2 \times 10 \times 900000}{4000^2} \):
\[ = \frac{18000000}{16000000} = 1.125 \]
Thus,
\[ R = \frac{900000}{1.125 - 1} = \frac{900000}{0.125} = 7200000 \, \text{m} = 7200 \, \text{km} \]

### Final Answer:
The radius of the planet is approximately \( 7200 \, \text{km} \).