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The escape velocity of 10g body from the...

The escape velocity of 10g body from the earth is 11.2 `km s^(-1)`. Ignoring air resistance, the escape velocity of 10 kg of the iron ball from the earth will be

A

`0.0112 kms^(-1)`

B

`0.112 kms^(-1)`

C

`11.2 kms^(-1)`

D

`0.56 kms^(-1)`

Text Solution

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The correct Answer is:
To solve the problem of finding the escape velocity of a 10 kg iron ball from the Earth, we can follow these steps: ### Step 1: Understand the Concept of Escape Velocity Escape velocity is the minimum speed needed for an object to break free from the gravitational attraction of a celestial body without any additional propulsion. It is given by the formula: \[ V_e = \sqrt{2gR} \] where: - \( V_e \) = escape velocity - \( g \) = acceleration due to gravity (approximately \( 9.8 \, \text{m/s}^2 \) on Earth) - \( R \) = radius of the Earth (approximately \( 6400 \, \text{km} = 6.4 \times 10^6 \, \text{m} \)) ### Step 2: Recognize that Escape Velocity is Independent of Mass It is important to note that escape velocity does not depend on the mass of the object. This means that whether the object is 10 g or 10 kg, the escape velocity remains the same. ### Step 3: Use the Given Escape Velocity The problem states that the escape velocity for a 10 g body from the Earth is \( 11.2 \, \text{km/s} \). Since escape velocity is independent of mass, we can conclude that: \[ V_e = 11.2 \, \text{km/s} \] ### Step 4: State the Final Answer Thus, the escape velocity of a 10 kg iron ball from the Earth is also: \[ \text{Escape Velocity} = 11.2 \, \text{km/s} \] ### Summary of the Solution The escape velocity of a 10 kg iron ball from the Earth is \( 11.2 \, \text{km/s} \), the same as that of a 10 g body. ---

To solve the problem of finding the escape velocity of a 10 kg iron ball from the Earth, we can follow these steps: ### Step 1: Understand the Concept of Escape Velocity Escape velocity is the minimum speed needed for an object to break free from the gravitational attraction of a celestial body without any additional propulsion. It is given by the formula: \[ V_e = \sqrt{2gR} \] ...
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