An unmanned space probe is to the thrown with such a velocity that it does not return to earth. It should be thrown witha)velocity ≥ (2GM/ R)1/2b)a great velocity but nothing more can be saidc)velocity >(2GM)1/2/ R necessarilyd)only the velocity = (2GM)1/2/RCorrect answer is option 'A'. Can you explain this answer?

Question

An unmanned space probe is to the thrown with such a velocity that it does not return to earth. It should be thrown witha)velocity &#8805; (2GM/ R)1/2b)a great velocity but nothing more can be saidc)velocity &#62;(2GM)1/2/ R necessarilyd)only the velocity = (2GM)1/2/RCorrect answer is option 'A'. Can you explain this answer?

Answer

It should be thrown with a velocity greater than or equal to escape velocity of the planet only then it will not return to Earth
Escape Velocity = √(2GM/R) = (2GM/R)^1/2
Hence A is the correct answer.

Answer

It should be thrown with a velocity greater than or equal to  escape velocity of the planet......then only it will not return the earth......escape velocity =√2GM/R=2GM^1/2/R......

Answer

No

Answer

Explanation:

The question asks for the velocity at which an unmanned space probe should be thrown in order for it to not return to Earth. The correct answer is option A, which states that the velocity should be (2GM/R)^(1/2).

Newton's Law of Universal Gravitation:
To understand why this is the correct answer, we need to first understand Newton's Law of Universal Gravitation. According to this law, every object with mass attracts every other object with mass with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers.

Escape Velocity:
In the context of this question, the concept of escape velocity is relevant. Escape velocity is the minimum velocity required for an object to escape the gravitational pull of a planet or celestial body. If an object is thrown with a velocity less than the escape velocity, it will eventually fall back to the planet due to gravity.

Deriving the Escape Velocity:
To derive the expression for escape velocity, we consider a probe of mass m being thrown from the surface of Earth with a velocity v. The force of gravity acting on the probe is given by F = (GMm)/R^2, where G is the gravitational constant, M is the mass of the Earth, and R is the radius of the Earth.

In order for the probe to escape the gravitational pull of the Earth, the force of gravity must be overcome. This can be achieved by giving the probe enough kinetic energy to counteract the gravitational force.

The kinetic energy of the probe is given by KE = (1/2)mv^2, and the potential energy due to gravity is given by PE = (-GMm)/R.

At the escape velocity, the kinetic energy is equal to the absolute value of the potential energy, so we have:

(1/2)mv^2 = (GMm)/R

Simplifying this equation, we find:

v^2 = (2GM)/R

Taking the square root of both sides, we get:

v = (2GM/R)^(1/2)

This is the expression for the escape velocity of an object near the surface of the Earth.

Conclusion:
Therefore, in order for an unmanned space probe to not return to Earth, it should be thrown with a velocity of (2GM/R)^(1/2), as stated in option A.

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