Problem 50 - A Complete Resource Book in Physics for JEE Main

Question

Suppose you are, as the header of a group of scientists working in NASA, sent to a planet named NSP2009 to study that planet you have following data with you. 1\. The planet NSP2009 is spherical in shape and of radius \(R\). 2\. The mass of the planet NSP2009 is M. If your team finds that the weight of any body inside the planet remains same as on its surface, what could be the possible value of mass density of the planet as a function of radial distance (where \(\rho_{0}\) is a content) \((R \geq>0)\) (A) \(\rho=\rho_{0} r^{0}\) (B) \(\rho=\rho_{0} r\) (C) \(\rho=\rho_{0} r^{-1}\) (D) \(\rho=\rho_{0} r^{-1 / 2}\)

Step-by-Step Solution

Answer

The possible mass density of the planet as a function of radial distance is \(\rho = \rho_{0} = \frac{3M}{4\pi R^3}\). The correct option is (A).

Step 1: Gravitational force inside and on the surface of the planet

Let's consider a body of mass m located at a distance r from the center of the planet (inside the planet). According to Newton's law of gravitation, the gravitational force acting on the body is given by: \[F = G \frac{M_{r}m}{r^2}\] where \(M_{r}\) is the mass of the planet contained within the radius r and G is the gravitational constant. On the surface of the planet (r = R), the gravitational force acting on the same body of mass m is: \[F_{s} = G \frac{Mm}{R^2}\] Since the gravitational force inside the planet remains the same as on its surface: \[F = F_{s}\]

Step 2: Express the mass of the planet within radius r

To express the mass of the planet within the radius r, we need to integrate the mass density function \(\rho_r\) over the volume of the sphere with radius r: \[M_{r} = \int_{0}^{r} \rho(\tilde{r}) 4\pi \tilde{r}^2 d\tilde{r}\]

Step 3: Set up an equation and solve for the mass density function

Now, we can set up an equation relating \(F\) and \(F_{s}\): \[G \frac{M_{r}m}{r^2} = G \frac{Mm}{R^2}\] Solving for \(M_{r}\): \[M_{r} = \frac{r^2}{R^2}M\] Now we can substitute the expression for \(M_{r}\) obtained from the integral: \[\int_{0}^{r} \rho(\tilde{r}) 4\pi \tilde{r}^2 d\tilde{r} = \frac{r^2}{R^2}M\] Now let's check each of the options for the mass density function and find the one that satisfies this equation. (A) \(\rho=\rho_{0} r^{0}\) If \(\rho = \rho_{0}\), then the integral and equation would be: \[\rho_{0} 4\pi \int_{0}^{r} \tilde{r}^2 d\tilde{r} = \frac{r^2}{R^2}M\] Solving the integral: \[\rho_{0} 4\pi \left[\frac{1}{3}\tilde{r}^3\right]_0^r = \frac{r^2}{R^2}M\] \(\Rightarrow \rho_{0} \frac{4\pi r^3}{3} = \frac{r^2}{R^2}M\) \(\Rightarrow \rho_{0} = \frac{3M}{4\pi R^3}\) Since this option results in a constant mass density throughout the planet, the weight inside the planet remains the same as on its surface, and this is the correct option. Therefore, the possible mass density of the planet as a function of radial distance is: \[\rho = \rho_{0} = \frac{3M}{4\pi R^3}\] The correct option is (A).

Key Concepts

Mass Density

Mass density is a fundamental concept in physics, representing how much mass is contained in a given volume of a substance. In simple terms, it tells us how "packed" the matter is in an object. For a spherical planet like NSP2009, mass density can change with radial distance from the center of the planet. It’s essential to express mass density mathematically to determine how mass is distributed across the planet. In our exercise, we explored different possible functions for mass density based on radial distance:

• Option (A): \( \rho = \rho_{0} \), indicating a constant mass density regardless of distance.
• Option (B): \( \rho = \rho_{0} r \), meaning mass density increases linearly with distance.
• Option (C): \( \rho = \rho_{0} r^{-1} \), where mass density decreases as we move away from the center.
• Option (D): \( \rho = \rho_{0} r^{-1/2} \), suggesting mass density decreases at a root-negative power of distance.

When mass density remains constant, it simplifies calculating gravitational forces within the planet, as demonstrated in Option (A). This option implies that mass is evenly distributed throughout the planet, making it the correct choice for this exercise.

Newton's Law of Gravitation

Newton's Law of Gravitation is a vital principle that describes the gravitational attraction between two masses. It states that every point mass attracts every other point mass in the universe with a force proportional to the product of their masses and inversely proportional to the square of the distance between their centers. This can be nicely encapsulated in the equation:\[ F = G \frac{M_{1}M_{2}}{r^2} \]Where:

• \( F \) is the gravitational force between the masses,
• \( G \) is the gravitational constant,
• \( M_{1} \) and \( M_{2} \) are the masses involved,
• \( r \) is the distance between the centers of the two masses.

In our context, Newton's Law of Gravitation helps determine how the gravitational force inside NSP2009 remains the same as on its surface. By equating the gravitational forces at different depths and at the planet's surface, we can deduce the planet's mass distribution, as illustrated in the exercise. Understanding this law is indispensable for exploring planetary physics and behavior of large bodies under mutual attraction.

Spherical Planet

A spherical planet like NSP2009 is identified by its symmetric shape where every point on the surface is equidistant from the center. This symmetry simplifies many calculations in physics, particularly those involving gravitational forces. The sphericity ensures that any variations in gravitational force are only due to changes in radial distance, not lateral distances. For a spherical planet, the total mass \( M \) and radius \( R \) can be used in calculations of gravitational forces and potential energy. The symmetry also implies that tools like integration over spherical volumes become plausible, as applied in determining mass density functions within the planet. Sphericity plays a critical role in permitting assumptions like constant mass density or interpreting gravitational behavior as experienced by our fictitious scientists on NSP2009.

Gravitational Constant

The gravitational constant, denoted as \( G \), is a crucial component of Newton's Law of Gravitation. It defines the strength of the gravitational force exerted between two bodies. Although its value is very small (\[ G \approx 6.674 \times 10^{-11} \ \,\text{N(m/kg)}^2 \]), it enables us to calculate the gravitational attractions that govern planetary motions and hold galaxies together. In the context of NSP2009, the gravitational constant allows us to compare forces inside and on the surface of the planet. It is essential for equating the expressions of gravitational force that lead to determining the required mass density function. An understanding of \( G \) provides valuable insights into how gravitational interactions scale with mass and distance, making it indispensable for astrophysics and cosmic studies.