Question

A particle is projected from this satellite radially outward relative to satellite with speed \(u\). Find the minimum value of \(u\) so that particle escape from the gravitational force of earth. (Mass of earth \(>>>\) mass of satellite \(>>>\) mass of particle) (1) \(\sqrt{\frac{G M}{R}}\) (2) \(\sqrt{\frac{G M}{2 R}}\) (3) \(\sqrt{\frac{2 G M}{R}}\) (4) \(2 \sqrt{\frac{G M}{R}}\)

Short answer

\( \sqrt{\frac{2GM}{R}} \)

Step-by-step solution

- Understand the concept of escape velocity

Escape velocity is the minimum speed needed for an object to escape from the gravitational influence of a celestial body without further propulsion. For Earth, the escape velocity from the surface is calculated with the formula: \[ v_e = \sqrt{\frac{2GM}{R}} \]

- Recall the general formula of escape velocity from any distance

The escape velocity from a distance \(r\) from the center of a celestial body of mass \(M\) is given by: \[ v_e = \sqrt{\frac{2GM}{r}} \]

- Apply the formula for escape velocity

Since the question involves a satellite orbiting the Earth, set the orbit radius as \(R\). Here, particle must escape the Earth's gravitational field from this orbit: \[ v_e = \sqrt{\frac{2GM}{R}} \]

- Match with the given options

Among the given options, the minimum value of \(u\) ensuring the particle escapes from Earth's gravitational force is: \[ \sqrt{\frac{2GM}{R}} \] Thus, the correct option is (3).

Key concepts

gravitational force

Gravitational force is the attractive force that exists between any two masses. It is described by Newton's law of universal gravitation, which states: \[ F = G\frac{m_1 m_2}{r^2} \] Where:

• F is the gravitational force between the masses.
• G is the gravitational constant ( \[ 6.67430 \times 10^{-11} \; m^3 \; kg^{-1} \; s^{-2} \]).
• m_1 and m_2 are the masses of the two objects.
• r is the distance between the centers of the two masses.

For any object near the Earth's surface, this force is what gives the object weight. The gravitational force keeps satellites in orbit and governs the motion of planets, moons, and other celestial bodies. Without it, the particle in our exercise wouldn't need an escape velocity to leave the Earth's influence.

satellite orbit

A satellite orbit is the path that a satellite follows around a celestial body like Earth. It is determined by a balance between gravitational force and the satellite's velocity. There are different types of orbits, such as low Earth orbit (LEO), geostationary orbit (GEO), and polar orbit, but in general, a satellite stays in orbit due to the following:

• The gravitational pull of the Earth which acts as a centripetal force.
• The satellite's tangential velocity which prevents it from falling straight down.

The speed required for maintaining an orbit just above the Earth's surface is known as orbital velocity, given by: \[v_o = \sqrt{\frac{GM}{R}} \] This is different from escape velocity which is higher because the satellite must overcome Earth's gravitational force entirely to leave its orbit. In our exercise, the particle needs a speed of \[\sqrt{\frac{2GM}{R}}\] to escape, not just to remain in orbit.

celestial mechanics

Celestial mechanics is the branch of astronomy that deals with the motions and gravitational forces of celestial objects. It involves the application of physics to understand how bodies like planets, moons, and satellites move under the influence of gravitational forces. Key concepts include:

• Kepler's Laws: Laws that describe the motion of planets around the Sun.
• Newton's Law of Gravitation: Explains how all objects with mass attract each other.
• Tsiolkovsky's Rocket Equation: Relates the motion of rockets with changing mass.

In our exercise, we specifically deal with the mechanics of how a particle can escape the Earth's gravity. This involves understanding how gravity decreases with distance and how much energy (and thus speed) the particle needs to break free from the gravitational pull. Through celestial mechanics, we derive the escape velocity formula used in our solution: \[v_e = \sqrt{\frac{2GM}{R}} \] This ensures that the particle has enough kinetic energy to reach a point where the gravitational pull of Earth becomes negligible.