Problem 697 - JEE Physics

Question

If the mass of earth is 80 times of that of a planet and diameter is double that of planet and ' \(\mathrm{g}\) ' on the earth is \(9.8 \mathrm{~ms}^{-2}\), then the value of \(\mathrm{g}^{\prime}\) on that planet is \(=\ldots \ldots \ldots \mathrm{ms}^{-2}\) (A) \(4.9\) (B) \(0.98\) (C) \(0.49\) (D) 49

Step-by-Step Solution

Answer

The value of $g'$ on the planet is \(4.9 \mathrm{~ms}^{-2}\). The correct choice is (A).

Step 1: Define the given quantities

We are given: 1. Mass of Earth is 80 times the mass of the planet: \(M_{Earth} = 80 M_{Planet}\) 2. Diameter of Earth is 2 times the diameter of the planet: \(D_{Earth} = 2D_{Planet}\) 3. g on Earth = \(9.8 \mathrm{~ms}^{-2}\)

Step 2: Find the relationship between g and mass

The formula for gravitational force is: \(F = \cfrac{GMm}{r^{2}}\) Where: F is the gravitational force between two masses, G is the gravitational constant (\(6.674 \times 10^{-11} \mathrm{Nm^{2}kg^{-2}}\)), M and m are the two masses, r is the distance between their centers. The force acting on an object due to gravity is the gravitational force, so using F = ma, \(ma = \cfrac{GMm}{r^{2}}\) Here, a is the acceleration due to gravity (g), so: \(g = \cfrac{GM}{r^{2}}\) The value of g depends on the mass of the planet and the distance from its center (which is half of the diameter, i.e., its radius).

Step 3: Solve for g' on the planet

Let g' be the acceleration due to gravity on the planet. Using the given data, we can rewrite the relationship found earlier: \(g' = \cfrac{G(80 M_{Planet})}{(2r_{planet})^{2}}\) Simplify the equation: \(g' = \cfrac{80 GM_{Planet}}{4r_{planet}^{2}}\) \(g' = \cfrac{20 GM_{Planet}}{r_{planet}^{2}}\) Now, we have the relationship \(g_{Earth} = \cfrac{GM_{Earth}}{r_{Earth}^{2}}\) and since \(g_{Earth} = 9.8 \mathrm{~ms}^{-2}\), we can plug the values. \(9.8 = \cfrac{G(80 M_{Planet})}{r_{Earth}^{2}}\) Dividing the two equations: \(\cfrac{g'}{9.8} = \cfrac{\cfrac{20 GM_{Planet}}{r_{planet}^{2}}}{\cfrac{80 GM_{Planet}}{r_{Earth}^{2}}}\) Cancel terms: \(\cfrac{g'}{9.8} = \cfrac{20}{80}\cfrac{r_{Earth}^{2}}{r_{planet}^{2}}\) Given relationship: \(D_{Earth} = 2D_{Planet}\), so \(r_{Earth} = 2 r_{Planet}\) \(\cfrac{g'}{9.8} = \cfrac{20}{80}\left(\cfrac{2r_{planet}}{r_{planet}}\right)^{2}\) \(\cfrac{g'}{9.8} = \cfrac{20}{80} (2^{2})\) \(\cfrac{g'}{9.8} = \cfrac{1}{2}\) Finally, solve for g': \(g' = 9.8 \times \cfrac{1}{2}\) \(g' = 4.9 \mathrm{~ms^{-2}}\)

Step 4: Compare with the given choices

The calculated value of g' on the planet is \(4.9 \mathrm{~ms}^{-2}\), which matches option (A). Therefore, the correct choice is (A) \(4.9 \mathrm{~ms}^{-2}\).

Key Concepts

Mass and Gravitational Acceleration

In the context of gravitational acceleration, mass is an essential factor to consider. The gravitational force that an object exerts depends on its mass. Larger masses exert more gravitational force. For example, the Earth has a substantial mass, which results in a gravitational acceleration of approximately \(9.8 \mathrm{~ms}^{-2}\). The formula for gravitational acceleration \(g\) is given by:\[ g = \frac{GM}{r^2} \]Here:

• \(G\) is a gravitational constant.
• \(M\) is the mass of the object, such as a planet.
• \(r\) is the distance from the center of the mass to the point of measurement.

When comparing two bodies like Earth and a planet, the gravitational acceleration experienced on a planet with a mass \(M_{Planet}\) much smaller than Earth's would obviously have a lower gravitational pull.

Radius and Its Effect

The radius of a planet significantly affects gravitational acceleration as it represents the distance over which the planet's gravity acts. The larger the radius, the weaker the gravitational acceleration at the surface because the force spreads over a larger area. The fundamental relation, again, is \(g = \cfrac{GM}{r^{2}}\), indicating that g is inversely proportional to the square of the radius (\(r^2\)). This means even a slight increase in the radius results in a significant decrease in gravitational acceleration. In our exercise, the diameter (and thus the radius) of Earth is given as twice that of the planet, which alters the value of \(r\) in the equation inversely, thereby affecting the gravitational acceleration greatly. This shows why understanding a planet's radius is key when we're investigating how strong its gravity will feel.

Gravitational Constant and Its Role

The gravitational constant \(G\) is a universal constant and is instrumental in the calculations involving gravity. Its value is \(6.674 \times 10^{-11} \mathrm{Nm^{2}kg^{-2}}\). It provides the framework that allows us to calculate the gravitational pull between two masses; without \(G\), we wouldn't be able to calculate gravitational forces accurately.\(G\) is vital in our equations:

• \(F = \cfrac{GMm}{r^{2}}\) for gravitational force,
• \(g = \cfrac{GM}{r^{2}}\) for gravitational acceleration.

These formulas show that \(G\) acts consistently across calculations, making it essential for understanding gravity not only on Earth but in celestial mechanics across different planets and systems.

Planetary Comparison

Comparing gravity on different planets involves looking at both mass and radius. With the exercise given, the problem highlights how changes in these quantities affect the gravity reading:

• The Earth is much more massive, having 80 times the mass of the planet compared.
• Its diameter is twice the planet's, which directly influences the gravitational pull felt on its surface.
• The acceleration due to gravity \(g'\) on the planet is calculated to be half of Earth's gravity value as a result of mass and radius variations.

This kind of analysis allows scientists to estimate not just gravitational forces but also how surface conditions, like weight that can be felt, would differ across celestial bodies. Such comparisons are foundational in astronomy and space exploration, helping us understand not just our planet but the wider universe.