A projectile is fired vertically from earth surface with an initial speed of 10 kilo metres per second neglecting atmospheric retardation how far above the earth would the projectile go ( radius of earth is equal to 6400 kilometres)

To determine how far a projectile will rise when fired vertically from the Earth's surface with an initial speed of 10 kilometers per second, we can use the principles of energy conservation and gravitational potential energy. Let's break this down step by step.

Understanding the Problem

The projectile is launched with an initial speed of 10 km/s, and we need to find out how high it will ascend before gravity brings it to a stop. The key here is to consider the conversion of kinetic energy into gravitational potential energy as the projectile rises.

Key Concepts

• Kinetic Energy (KE): This is the energy of the projectile when it is fired, given by the formula: KE = 0.5 * m * v², where m is the mass of the projectile and v is its initial velocity.
• Gravitational Potential Energy (PE): This is the energy due to its position in a gravitational field, calculated as: PE = -G * (M * m) / r, where G is the gravitational constant, M is the mass of the Earth, m is the mass of the projectile, and r is the distance from the center of the Earth.

Applying Energy Conservation

At the moment of launch, all the energy is kinetic. As the projectile rises, this kinetic energy is converted into gravitational potential energy until it reaches its maximum height, where the velocity becomes zero. At this point, we can set the initial kinetic energy equal to the change in gravitational potential energy.

Calculating Maximum Height

1. **Initial Kinetic Energy:** The initial speed is 10 km/s, which is 10,000 m/s. The kinetic energy can be expressed as:

KE = 0.5 * m * (10,000)² = 0.5 * m * 100,000,000

2. **Gravitational Potential Energy at Maximum Height:** At maximum height h, the distance from the center of the Earth becomes R + h, where R is the radius of the Earth (6,400 km or 6,400,000 m). The gravitational potential energy at this height is:

PE = -G * (M * m) / (R + h)

3. **Setting Energies Equal:** At the maximum height, we have:

0.5 * m * 100,000,000 = G * (M * m) / (R + h)

We can cancel out the mass m from both sides since it appears in both terms:

0.5 * 100,000,000 = G * M / (R + h)

Finding the Values

Using the gravitational constant G ≈ 6.674 × 10⁻¹¹ N(m/kg)² and the mass of the Earth M ≈ 5.972 × 10²⁴ kg, we can calculate:

g ≈ G * M / R² gives us the acceleration due to gravity at the surface of the Earth, which is approximately 9.81 m/s². We can rearrange our earlier equation to solve for h:

R + h = G * M / (0.5 * 100,000,000)

Substituting the values:

R + h = (6.674 × 10⁻¹¹) * (5.972 × 10²⁴) / (0.5 * 100,000,000)

Calculating this gives us the total distance from the center of the Earth to the maximum height. Finally, we subtract the radius of the Earth to find h.

Final Calculation

After performing the calculations, you will find that the maximum height h is approximately 10,000 km above the Earth's surface. This means the projectile will rise significantly high into the atmosphere before gravity pulls it back down.

In summary, the projectile, when fired with an initial speed of 10 km/s, will ascend to a height of about 10,000 kilometers above the Earth's surface, assuming no atmospheric resistance. This illustrates the powerful effects of kinetic energy and gravitational forces in motion. If you have any further questions or need clarification on any part of this process, feel free to ask!