Problem 833 - JEE Physics

Question

On the surface of earth acceleration due to gravity is \(\mathrm{g}\) and gravitational potential is \(\mathrm{V}\) match the followingTable - 1 Table \(-2\) (A) At height \(\mathrm{h}=\mathrm{R}\) value of \(\mathrm{g}\) (P) decrease by a factor \((1 / 4)\) (B) At depth \(\mathrm{h}=(\mathrm{R} / 2)\) (Q) decrease by a factor \((1 / 2)\) (C) At height \(\mathrm{h}=\mathrm{R}\) value of \(\mathrm{v}\) (R) increase by a factor \((11 / 8)\) (D) At depth \(\mathrm{h}=(\mathrm{R} / 2)\) value of \(\mathrm{v}\) (S) increase by a factor 2 (T) None

Step-by-Step Solution

Answer

The final matching for the given options is: A -> P: At height \(h=R\), the value of g decreases by a factor of \(\frac{1}{4}\). B -> Q: At depth \(h=\frac{R}{2}\), the value of g decreases by a factor of \(\frac{1}{2}\). C -> T: At height \(h=R\), the value of V doesn't match any given option. D -> T: At depth \(h=\frac{R}{2}\), the value of V doesn't match any given option.

Step 1: 1. Evaluate the change in gravitational acceleration at height h = R

To determine the change in gravitational acceleration at height h = R, we can use the formula: \(g'=g\frac{R^2}{(R+h)^2}\) Where: \(g'\) = gravitational acceleration at height h \(g\) = gravitational acceleration on Earth's surface \(R\) = radius of Earth \(h\) = height above Earth's surface Given, \(h=R\), so we have: \(g'=g\frac{R^2}{(R+R)^2} = g\frac{R^2}{(2R)^2} = g\frac{1}{4}\) The gravitational acceleration at height \(h=R\) decreases by a factor of \(\frac{1}{4}\). So, the correct match for option A is P.

Step 2: 2. Evaluate the change in gravitational acceleration at depth h = R/2

To determine the change in gravitational acceleration at depth h = R/2, we can use the formula: \(g''=g(1-\frac{h}{R})\) Where: \(g''\) = gravitational acceleration at depth h \(g\) = gravitational acceleration on Earth's surface \(R\) = radius of Earth \(h\) = depth below Earth's surface Given, \(h=\frac{R}{2}\), so we have: \(g''=g(1-\frac{R/2}{R}) = g(1-\frac{1}{2}) = g\frac{1}{2}\) The gravitational acceleration at depth \(h=R/2\) decreases by a factor of \(\frac{1}{2}\). So, the correct match for option B is Q.

Step 3: 3. Evaluate the change in gravitational potential at height h = R

To determine the change in gravitational potential at height h = R, we can use the formula: \(V'=-\frac{GM}{R+h}\) Where: \(V'\) = gravitational potential at height h \(G\) = gravitational constant \(M\) = mass of Earth \(R\) = radius of Earth \(h\) = height above Earth's surface Given, \(h=R\), so we have: \(V'=-\frac{GM}{R+R} = -\frac{GM}{2R}\) Initially, the gravitational potential on Earth's surface is: \(V = -\frac{GM}{R}\) Now let's find the factor by which the gravitational potential increases: \(\frac{V'}{V} = \frac{-GM/2R}{-GM/R} = \frac{1}{2}\) This means the gravitational potential at height \(h=R\) increases by a factor of \(\frac{1}{2}\). However, there is no matching option for this case, so the correct match for option C is T (None).

Step 4: 4. Evaluate the change in gravitational potential at depth h = R/2

To determine the change in gravitational potential at depth h = R/2, we can use the formula: \(V''=-GM\frac{3}{2R}\) Where: \(V''\) = gravitational potential at depth h \(G\) = gravitational constant \(M\) = mass of Earth \(R\) = radius of Earth \(h\) = depth below Earth's surface Given, \(h=\frac{R}{2}\), so we have: \(V''=-GM\frac{3}{2R}\) Now let's find the factor by which the gravitational potential increases: \(\frac{V''}{V} = \frac{-GM(3/2R)}{-GM/R} = \frac{3}{2}\) This means the gravitational potential at depth \(h=R/2\) increases by a factor of \(\frac{3}{2}\). However, there is no matching option for this case, so the correct match for option D is T (None). So, the final matching is: A -> P B -> Q C -> T D -> T

Key Concepts

Acceleration Due to Gravity

On Earth, the acceleration due to gravity is often denoted by the symbol \(g\) and is approximately 9.8 m/s² at the surface. This constant influences how quickly objects fall to the ground due to Earth's gravitational pull. It's an important factor to consider when studying motion and physics.

• At heights or altitudes above the Earth's surface, such as \(h = R\), where \(R\) is the radius of Earth, gravity decreases. The new gravitational force can be calculated using the formula \(g' = g \frac{R^2}{(R+h)^2}\). For example, at \(h = R\), it results in a decreased value of gravitational force by a factor of \(\frac{1}{4}\).
• Within the Earth, at depths, such as \(h = R/2\), gravity also diminishes, calculated as \(g'' = g (1-\frac{h}{R})\). At this depth, acceleration due to gravity decreases by a factor of \(\frac{1}{2}\).

Gravitational acceleration varies depending on your position relative to Earth's surface, whether you are above it in the atmosphere, or beneath it underground.

Gravitational Potential

Gravitational potential, denoted as \( V \), is a measure of the work done in moving a unit mass from a reference point (often infinity) to a particular point in space. In simpler terms, it's a way to quantify the energy needed to bring an object from a distant point to within Earth's gravitational influence.

• At the Earth's surface, gravitational potential is represented as \( V = -\frac{GM}{R} \), where \( G \) is the gravitational constant, and \( M \) is Earth's mass.
• When you move to a height \( h = R \), the gravitational potential changes, calculated using the formula \( V' = -\frac{GM}{R+h} \). This results in a change, showing an increase by a factor potentially desired for specific physical scenarios.
• Going deeper, such as to a depth \( h = R/2 \), transforms gravitational potential as calculated with \( V'' = -GM\frac{3}{2R} \). Again, noting how gravitational potential increases, but lacks a specific option in the provided context.

Understanding gravitational potential helps explain how energy transfers in the gravitational field, crucial for planetary science and astrophysics.

Radius of Earth

The radius of the Earth, symbolized as \(R\), is an essential constant in gravitational calculations, providing an average distance from the Earth's core to its surface. It's approximately 6,371 kilometers. This measurement plays a crucial role in determining values such as acceleration due to gravity and gravitational potential.

• The radius is used in formulas like \(g' = g \frac{R^2}{(R+h)^2}\), which calculates the change in gravity at a certain height \( h \).
• Also integral in expressions such as \(V = -\frac{GM}{R}\), revealing how gravitational potential varies depending on distance from Earth's center.

The radius significantly influences various physical phenomena and engineering calculations, like satellite trajectories and Earth's shape, considering both revolutions and local gravity variations.