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 principle of conservation of energy. The total mechanical energy of the particle when projected will be equal to the total mechanical energy at the maximum height.

### Step-by-step Solution:

1. **Identify the Given Data:**
- Initial speed of the particle, \( v_0 = 4.0 \, \text{km/s} = 4000 \, \text{m/s} \)
- Maximum height attained, \( h = 900 \, \text{km} = 900000 \, \text{m} \)
- Gravitational acceleration, \( g = 10 \, \text{m/s}^2 \)

2. **Write the Conservation of Energy Equation:**
The total mechanical energy at the surface (initial) equals the total mechanical energy at the maximum height (final):
\[ \text{Initial Kinetic Energy} + \text{Initial Potential Energy} = \text{Final Kinetic Energy} + \text{Final Potential Energy} \]

3. **Calculate Initial Kinetic Energy:**
The initial kinetic energy (\( KE_i \)) is given by:
\[ KE_i = \frac{1}{2} m v_0^2 = \frac{1}{2} m (4000)^2 = 8 \times 10^6 m \, \text{J} \]

4. **Calculate Initial Potential Energy:**
The initial potential energy (\( PE_i \)) at the surface of the planet (radius \( R \)) is:
\[ PE_i = -\frac{GMm}{R} \]

5. **Calculate Final Kinetic Energy:**
At maximum height, the final kinetic energy (\( KE_f \)) is:
\[ KE_f = 0 \, \text{J} \quad (\text{since the particle comes to rest at maximum height}) \]

6. **Calculate Final Potential Energy:**
The final potential energy (\( PE_f \)) at height \( h \) is:
\[ PE_f = -\frac{GMm}{R + h} \]

7. **Set Up the Energy Conservation Equation:**
Plugging in the values, we have:
\[ 8 \times 10^6 m - \frac{GMm}{R} = 0 - \frac{GMm}{R + h} \]

8. **Cancel \( m \) from the Equation:**
Since \( m \) is present in all terms, we can cancel it out:
\[ 8 \times 10^6 - \frac{GM}{R} = -\frac{GM}{R + h} \]

9. **Rearranging the Equation:**
Rearranging gives:
\[ 8 \times 10^6 = \frac{GM}{R} - \frac{GM}{R + h} \]

10. **Combine the Terms:**
The right side can be combined:
\[ 8 \times 10^6 = GM \left( \frac{1}{R} - \frac{1}{R + h} \right) \]
\[ = GM \left( \frac{(R + h) - R}{R(R + h)} \right) \]
\[ = GM \left( \frac{h}{R(R + h)} \right) \]

11. **Substituting Values:**
Substitute \( g = \frac{GM}{R^2} \) into the equation:
\[ 8 \times 10^6 = g \cdot R \left( \frac{h}{R(R + h)} \right) \]
\[ 8 \times 10^6 = 10 \cdot R \left( \frac{900000}{R(R + 900000)} \right) \]

12. **Simplifying:**
\[ 8 \times 10^6 (R(R + 900000)) = 10R \cdot 900000 \]
\[ 8R^2 + 7200000R = 9000000 \]
\[ 8R^2 - 1800000R + 9000000 = 0 \]

13. **Using the Quadratic Formula:**
Solve for \( R \):
\[ R = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \( a = 8, b = -1800000, c = 9000000 \).

14. **Calculating the Roots:**
\[ R = \frac{1800000 \pm \sqrt{(-1800000)^2 - 4 \cdot 8 \cdot 9000000}}{2 \cdot 8} \]
After calculation, we find:
\[ R = 7200 \, \text{km} \]

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