Problem 727 - JEE Physics

Question

A body is projected vertically upwards from the surtace of a planet of radius \(\mathrm{R}\) with a velocity equal to half the escape velocity for that planet. The maximum height attained by the body is \(\ldots \ldots \ldots \ldots\) (A) \((\mathrm{R} / 3)\) (B) \((\mathrm{R} / 2)\) (C) \((\mathrm{R} / 4)\) (D) \((\mathrm{R} / 5)\)

Step-by-Step Solution

Answer

The maximum height attained by the body is \(H = R\), which corresponds to option (B).

Step 1: Write down the known parameters

The radius of the planet is R. The initial velocity (v0) is equal to half the escape velocity, which is given by the formula: \(v_e = \sqrt{\frac{2GM}{R}}\), where G is the gravitational constant, and M is the mass of the planet. So, the initial velocity v0 is: \(v_0 = \frac{1}{2}v_e = \frac{1}{2}\sqrt{\frac{2GM}{R}}\)

Step 2: Apply the conservation of mechanical energy

We can write the total mechanical energy of the system (kinetic energy + gravitational potential energy) at launch (initial) and at the maximum height (final). Initially, the projectile has kinetic energy and potential energy due to its position on the planet's surface: \[E_{initial} = \frac{1}{2}mv_0^2 - \frac{GMm}{R}\] At the maximum height (H), the projectile's velocity is 0 (it momentarily stops moving upwards), so its kinetic energy is 0. However, it has potential energy due to its position at height H: \[E_{final} = -\frac{GMm}{R+H}\] Since the mechanical energy is conserved, we can write: \[E_{initial} = E_{final}\]

Step 3: Solve for the maximum height H

From the energy conservation equation, we get: \[\frac{1}{2}mv_0^2 - \frac{GMm}{R} = -\frac{GMm}{R+H}\] Substitute the initial velocity v0 into the equation and solve for the unknown height H: \[\frac{1}{2}m\left(\frac{1}{2}\sqrt{\frac{2GM}{R}}\right)^2 - \frac{GMm}{R} = -\frac{GMm}{R+H}\] Simplify and solve for H: \[\frac{1}{4}\frac{2GMm}{R} - \frac{GMm}{R} = -\frac{GMm}{R+H}\] \[-\frac{1}{2}\frac{GMm}{R} = -\frac{GMm}{R+H}\] Cancel out the negative signs and GMm, and cross-multiply: \[\frac{1}{2R}(R+H)=1\] Solve for H: \[H = R\left(\frac{1}{\frac{1}{2}} - 1\right) = R\left(2-1\right) = R\] Therefore, the maximum height attained by the body is \(H = R\), which corresponds to option (B).

Key Concepts

Escape Velocity

Escape velocity is the minimum speed an object needs to break free from a planet's gravitational pull without any further propulsion. Think of it like the rocket speed needed to go off into space. The formula for escape velocity \[ v_e = \sqrt{\frac{2GM}{R}} \]- **G** is the gravitational constant.- **M** is the mass of the planet.- **R** is the radius of the planet.If the body’s velocity is less than the escape velocity, it won't reach outer space; it might fall back instead. In the exercise, the body is launched with only half of the escape velocity. This means it won't leave the planet entirely, but instead, it will reach a maximum height before falling back. This helps us understand how much energy and speed are needed to escape a planet's grasp.

Conservation of Mechanical Energy

The principle of conservation of mechanical energy states that the total mechanical energy remains constant, unless acted upon by external forces. The mechanical energy is the sum of kinetic energy (energy of motion) and potential energy (stored energy due to position). For projectile motion near a planet: \[ E_{\text{initial}} = \frac{1}{2}mv_0^2 - \frac{GMm}{R} \]- **\(m\)** is the mass of the body.- **\(v_0\)** is the initial velocity, here half the escape velocity.At the maximum height, kinetic energy is zero because the body stops momentarily. Thus, only potential energy is considered.\[ E_{\text{final}} = -\frac{GMm}{R+H} \]Conservation of mechanical energy means:\[ E_{\text{initial}} = E_{\text{final}} \]By equating and simplifying these equations, we find the maximum height, helping us understand energy distribution in projectile motion.

Gravitational Potential Energy

Gravitational potential energy (GPE) is the energy an object possesses due to its position in a gravitational field. It increases as an object moves further from the center of a massive body, like a planet. The formula for calculating GPE is:\[ U = -\frac{GMm}{r} \]- **\(U\)** is gravitational potential energy.- **\(r\)** is the distance from the center of the planet.Initially, when the body is on the surface, \[ U = -\frac{GMm}{R} \]When the body reaches its maximum height \[ U = -\frac{GMm}{R+H} \]By understanding how GPE changes with height, we can better grasp why a body projected upwards with a certain velocity will reach a particular maximum height before descending. This concept is key in calculating how high an object can go based on its initial speed and the mass and radius of the planet.