Problem 737 - JEE Physics

Question

The escape velocity of an object from the earth depends upon the mass of earth (M), its mean density ( \(p\) ), its radius (R) and gravitational constant (G), thus the formula for escape velocity is (A) \(U=\mathrm{R} \sqrt{[}(8 \pi / 3) \mathrm{Gp}]\) (C) \(\mathrm{U}=\sqrt{(2 \mathrm{GMR})}\) (D) \(U=\sqrt{\left[(2 \mathrm{GMR}) / \mathrm{R}^{2}\right]}\)

Step-by-Step Solution

Answer

The correct formula for escape velocity is (C) \(U=\sqrt{2 \mathrm{GMR}}\).

Step 1: Establish mass, density, and volume relationship

To find the correct formula, we first need a relationship between mass, density, and volume. We know that mass (M) equals density (ρ) times volume (V). Therefore, we have: \[M = \rho V\] Since Earth can be approximated as a sphere, we'll use this to compute the volume of Earth: \[V = \frac{4}{3} \pi R^{3}\] where R is the radius of Earth.

Step 2: Calculate gravitational force

The gravitational force (F) between two bodies with masses (M1 and M2) and a distance R separating them is given by: \[F = G\frac{M_{1}M_{2}}{R^{2}}\] In our case, M1 is the mass of Earth (M) and M2 is the mass of the object trying to escape Earth. We can rewrite the equation as: \[F = G\frac{Mm}{R^{2}}\]

Step 3: Equate gravitational force to the work done

Escape velocity is defined as the minimum velocity an object needs to be propelled to escape the gravitational pull of a massive body, like Earth, without further propulsion. The work done against gravity must be equal to the gravitational force. Work done (W) can be given by: \[W = Fd\] where d is the distance traveled (in our case, it is equal to the radius R of Earth). We can rewrite the equation as: \[W = GMm / R\]

Step 4: Determine kinetic energy and escape velocity

The work done (W) is equal to the kinetic energy (K.E.) of the object which is trying to escape Earth's gravity, given by: \[K.E. = \frac{1}{2}mv^{2}\] where m is the mass of the object and v is its velocity. Set the work done equal to the kinetic energy: \[GMm / R = \frac{1}{2}mv^{2}\] where v represents the escape velocity.

Step 5: Solve for escape velocity

Next, we solve for v (escape velocity) from the equation derived in Step 4. First, divide both sides by the mass of the object (m): \[\frac{GM}{R} = \frac{1}{2}v^{2}\] Then, multiply both sides by 2 to eliminate the 1/2: \[2\frac{GM}{R} = v^{2}\] Finally, take the square root of both sides: \[v = \sqrt{2\frac{GM}{R}}\] Comparing this result to the given options, we find that the correct formula for escape velocity is: (C) \(U=\sqrt{2 \mathrm{GMR}}\)

Key Concepts

Gravitational Constant

The gravitational constant, often symbolized by \(G\), is a crucial part of understanding gravitational force and escape velocity. It is a measure of the strength of gravity. Specifically, it quantifies the force of attraction between two bodies with mass. - The gravitational constant is valued at approximately \(6.674 imes 10^{-11} \text{Nm}^2/\text{kg}^2\).- This value is a universal constant; it does not change regardless of the masses or the distance involved.
Gravitational constant plays a key role when calculating gravitational forces between two objects using the formula: \[ F = G\frac{M_1M_2}{R^2} \]Understanding \(G\) helps grasp how objects like planets and moons pull on other bodies in space, affecting orbits and paths. This constant is also crucial in determining escape velocity, as it directly relates to how much velocity an object needs to overcome Earth's gravitational pull.

Mass-Density Relationship

The relationship between mass, density, and volume is central to understanding how celestial bodies like Earth exert gravitational force. Mass \(M\) is determined by the density \(\rho\) of a substance and its volume \(V\). The mathematical relationship is expressed as: \[ M = \rho V \] - Density \(\rho\) is defined as mass per unit volume and tells us how compact a material is.- Volume, in the case of spherical bodies like the Earth, can be found using the formula: \[ V = \frac{4}{3}\pi R^3 \]where \(R\) is the radius of the sphere.
This relationship helps explain why bodies with high density can exert strong gravitational forces despite possibly being small in size. This interplay between mass and density is pivotal when calculating escape velocity, as it helps to determine how much force must be overcome to leave the Earth's gravitational field.

Gravitational Force

Gravitational force is the attractive force between two masses. This concept is essential to understanding why objects need a certain velocity to break away from a planet's gravitational pull. According to Newton's law of universal gravitation, the gravitational force \(F\) between two masses \(M_1\) and \(M_2\), separated by a distance \(R\), is calculated with: \[ F = G\frac{M_1M_2}{R^2} \] - This equation shows that gravitational force is directly proportional to the product of the two masses, and inversely proportional to the square of the distance between them.- Therefore, the larger the masses or the closer they are to each other, the stronger the gravitational force.
Recognizing these aspects of gravitational force is indispensable for understanding how escape velocity is formulated. It illustrates the gravitational pull an object must overcome to journey from the surface of a planet into space.

Kinetic Energy

Kinetic energy is the energy an object possesses due to its motion. For an object attempting to escape Earth's gravitational pull, understanding kinetic energy is necessary. The formula to calculate kinetic energy \(K.E.\) is: \[ K.E. = \frac{1}{2}mv^2 \] - Here, \(m\) is the mass of the object, and \(v\) is its velocity.- Kinetic energy increases with the square of the velocity, highlighting the energy required for higher escape velocities.
This concept is directly linked with the work done against gravitational force. For an object to escape Earth's gravity, the work done by its kinetic energy must equal the gravitational potential energy holding it down. Solving this equation gives the famous escape velocity formula, demonstrating the energy balance needed to overcome gravitational forces.