Problem 749 - JEE Physics

Question

The acceleration due to gravity on a planet is same as that on earth and its radius is four times that of earth. What will be the value of escape velocity on that planet if it is \(\mathrm{V}_{\mathrm{e}}\) on the earth (A) \(\mathrm{V}_{\mathrm{e}}\) (B) \(2 \mathrm{~V}_{\mathrm{e}}\) (C) \(4 \mathrm{~V}_{\mathrm{e}}\) (D) \(\mathrm{V}_{\mathrm{e}} / 2\)

Step-by-Step Solution

Answer

The escape velocity of the given planet is half of the escape velocity of Earth. Thus, the answer is (D) \(V_{e_{planet}} = \frac{V_{e_{Earth}}}{2}\).

Step 1: List the given information

The acceleration due to gravity on the planet is same as that on Earth: \(g_{planet} = g_{Earth}\) The radius of the planet is four times the radius of Earth: \(R_{planet} = 4 * R_{Earth}\)

Step 2: Write down the escape velocity formula

The escape velocity (\(V_e\)) is calculated using the formula: \[V_e = \sqrt{\frac{2GM}{R}}\] where - \(V_e\) is the escape velocity - \(G\) is the gravitational constant - \(M\) is the mass of the planet - \(R\) is the radius of the planet

Step 3: Express the mass of the planet in terms of Earth

The mass of the planet can be expressed in terms of Earth using the provided information about the gravity and radius of the planet. Since acceleration due to gravity on both planets is equal: \[g_{Earth} = g_{planet}\] Also, recall that \[g = \frac{GM}{R^2}\] So, for Earth and the given planet: \[\frac{G M_{Earth}}{R_{Earth}^2} = \frac{G M_{planet}}{R_{planet}^2}\] We know that \(R_{planet} = 4 R_{Earth}\). So, we can rewrite the equation: \[\frac{M_{Earth}}{R_{Earth}^2} = \frac{M_{planet}}{(4R_{Earth})^2}\] Solve the equation for the mass of the planet (\(M_{planet}\)): \[M_{planet} = \frac{1}{16} M_{Earth}\]

Step 4: Calculate the escape velocity of the given planet

Now we have enough information to calculate the escape velocity of the given planet. Substitute the values of \(M_{planet}\) and \(R_{planet}\) into the escape velocity formula: \[V_{e_{planet}} = \sqrt{\frac{2G}{16}M_{Earth} \div 4R_{Earth}}\] Simplifying the expression, we get: \[V_{e_{planet}} = \sqrt{\frac{G M_{Earth}}{32R_{Earth}}}\]

Step 5: Compare the escape velocity to Earth's escape velocity

Now, let's find the escape velocity of Earth: \[V_{e_{Earth}} = \sqrt{\frac{2GM_{Earth}}{R_{Earth}}}\] Now, divide \(V_{e_{planet}}\) by \(V_{e_{Earth}}\): \[\frac{V_{e_{planet}}}{V_{e_{Earth}}} = \frac{\sqrt{\frac{G M_{Earth}}{32R_{Earth}}}}{\sqrt{\frac{2GM_{Earth}}{R_{Earth}}}}\] Simplifying the expression, we find: \[\frac{V_{e_{planet}}}{V_{e_{Earth}}} = \frac{1}{2}\] Which indicates that the escape velocity of the given planet is half of the escape velocity of Earth. Thus, the answer is: \[\boxed{(D)~ V_{e_{planet}} = \frac{V_{e_{Earth}}}{2}}.\]

Key Concepts

Acceleration Due to Gravity

The acceleration due to gravity is the force that attracts a body towards the center of a planet. On Earth, this value is approximately 9.8 m/s². In the given problem, the acceleration due to gravity on the mystery planet is stated to be the same as on Earth. What does this mean for us? It means that both the Earth and our mystery planet have a gravitational pull that affects objects in the same way. This equivalence forms a key part of understanding the larger picture, especially when considering how escape velocity is calculated. Remember, this value relies on both the planet's mass and radius, linking us to our next important concept about gravitational relationships.

Gravitational Constant

The gravitational constant, denoted as G , is a crucial component in the fundamental equation of universal gravitation. It appears in the equation used to calculate escape velocity: \[ V_e = \sqrt{\frac{2GM}{R}} \] where G is assumed to have a constant value everywhere in the universe. It links the mass of two bodies with the force of attraction between them. In escape velocity scenarios, G helps us understand how powerful this attraction is when mass and radius of the body (or planet) are taken into account. This constant ensures that whether we're considering Earth or our new planet, the fundamental physics remain the same, which simplifies our calculations.

Planetary Radius

The radius of a planet is how far it reaches out from its center to its surface. In the exercise, the given planet's radius is four times that of Earth ( \(R_{planet} = 4 \times R_{Earth} \) ). The radius directly influences escape velocity, as seen in the formula: \[ V_e = \sqrt{\frac{2GM}{R}} \] A larger radius typically means a lower escape velocity, assuming mass doesn't increase at the same rate. Thus, a planet with a larger radius but the same gravitational pull indicates a unique mass distribution and influences how objects can leave that planet's gravitational field. Understanding this allows us to reconsider our initial perceptions of how massive objects behave under different planetary conditions.

Mass of the Planet

The mass of a planet, contributing significantly to its gravitational pull, plays a central role in determining escape velocity. It is evident from the equation \( g = \frac{GM}{R^2} \) , showing how gravity ( g ) is affected by both mass ( M ) and radius ( R ). For our mysterious planet, we derived \( M_{planet} = \frac{1}{16} M_{Earth} \) due to its altered radius and the same gravitational pull as Earth. This drastic drop in mass despite the same effective gravity is due to the increase in radius. Such a change underscores the complex relationship between mass, radius, and gravitational effects, essential for calculating attributes like escape velocity efficiently and accurately.