I wasn't going to post an answer to this, but the other answers seem very insufficient to me and may give the wrong idea. The work required to lift an object off the ground is not affected by velocity directly. The work is proportional to the product of the force acting on the object, and it's displacement in the direction of that force. This means that if you are applying the same force over the same distance, it takes the same amount of work, regardless of the velocity. The velocity is relevant to the power required; which is energy per unit time. The faster you want to travel the displacement, the more power is required, so you need to deliver the same amount of energy in a decreased time frame. You talk about overcoming gravity in the comments, and I will try to clear up confusion there as well. If gravity increases, to overcome gravity you would need a greater minimum force to overcome it's effects. Because of this, an increase in gravity will lead to an increase in required force for the same results, and if that force increases, the required work would increase as well. So to reiterate again, velocity has no direct bearing on the applied work. It will affect the required power, but the amount of work required to move the specified distance only depends on the displacement and the applied force; not upon how quickly it travels the displacement.
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The correct answer is option 2) i.e. R/2 CONCEPT: - Law of Universal Gravitation: It states that all objects attract each other with a force that is proportional to the masses of two objects and inversely proportional to the square of the distance that separates their centres.
\(F = \frac{Gm_1m_2}{R^2}\) Where m1 and m2 are the mass of two objects, G is the gravitational constant and R is the distance between their centres. - Gravitational Potential Energy: It is the energy possessed by a body at a certain point when work is done by the force of gravity in bringing the object from infinity to that point.
- The
**gravitational potential energy**at a point above the Earth's surface is given by
- The
Where M is the mass of earth, m is the mass of the object and Re is the radius of the earth. CALCULATION: - From Newton's second law, we know that F = ma = mg
Given that: Increase in potential energy, \(ΔU =\frac{1}{3}gmR\) Substituting the value of g in ΔU, \(⇒ ΔU =\frac{1}{3}\frac{GM}{R^2}mR = \frac{GMm}{3R}\) The object is raised to a height H from the surface of the earth ⇒ distance R2 = R + H At the surface of the earth ⇒ distance R1 = R Therefore the \(\Rightarrow ΔU = -\frac{GMm}{R\ +\ H} +\frac{GMm}{R}\) \(\Rightarrow \frac{GMm}{3R}= -\frac{GMm}{R\ +\ H} +\frac{GMm}{R}\) \(\Rightarrow \frac{1}{R\ +\ H} = \frac{1}{R} -\frac{1}{3R}\) \(\Rightarrow \frac{1}{R\ +\ H} = \frac{2}{3R}\) \(\Rightarrow H = \frac{R}{2}\)
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