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If a projectile lifts off from the surface of the Earth with a speed of 11.2 km.s–1, then it can escape from the Earth’s gravitational field completely. This is called the escape velocity. If the radius of the Earth were 2 times larger and the mass 8 times larger, then the escape velocity (in km.s–1) would be
- a)5.6
- b)11.2
- c)22.4
- d)44.8
Correct answer is option 'C'. Can you explain this answer?
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If a projectile lifts off from the surface of the Earth with a speed o...
To understand why the escape velocity would be 22.4 km/s if the radius of the Earth were 2 times larger and the mass 8 times larger, let's break down the concept step by step.
1. Escape Velocity:
Escape velocity is the minimum velocity an object needs to escape the gravitational pull of a celestial body, such as the Earth. It is the speed at which the kinetic energy of an object is equal to or greater than the gravitational potential energy.
2. Calculation of Escape Velocity:
The formula for escape velocity is given by:
Ve = √(2GM/R)
Where:
- Ve is the escape velocity
- G is the gravitational constant (approximately 6.67430 × 10^-11 m^3 kg^-1 s^-2)
- M is the mass of the celestial body (in this case, the Earth)
- R is the radius of the celestial body (in this case, the Earth)
3. Effect of Increased Radius and Mass:
If the radius of the Earth is 2 times larger, the new radius (R') would be 2R. Similarly, if the mass of the Earth is 8 times larger, the new mass (M') would be 8M.
4. Calculation of New Escape Velocity:
Using the formula for escape velocity, we can calculate the new escape velocity (Ve'):
Ve' = √(2G(M')/(R'))
Substituting the values of R' and M' in terms of R and M:
Ve' = √(2G(8M)/(2R))
Simplifying the equation:
Ve' = √(8GM/R)
Since Ve' = 2Ve (as per the given condition), we have:
2Ve = √(8GM/R)
Squaring both sides:
4Ve^2 = 8GM/R
Simplifying further:
Ve^2 = 2GM/R
Taking square root on both sides:
Ve = √(2GM/R)
Comparing this with the original formula for escape velocity, we can see that the new escape velocity (Ve') is equal to the square root of 2 times the original escape velocity (Ve).
So, if the radius of the Earth were 2 times larger and the mass 8 times larger, the new escape velocity (Ve') would be 22.4 km/s, which is 2 times the original escape velocity of 11.2 km/s.
Therefore, the correct answer is option 'C' (22.4 km/s).
1. Escape Velocity:
Escape velocity is the minimum velocity an object needs to escape the gravitational pull of a celestial body, such as the Earth. It is the speed at which the kinetic energy of an object is equal to or greater than the gravitational potential energy.
2. Calculation of Escape Velocity:
The formula for escape velocity is given by:
Ve = √(2GM/R)
Where:
- Ve is the escape velocity
- G is the gravitational constant (approximately 6.67430 × 10^-11 m^3 kg^-1 s^-2)
- M is the mass of the celestial body (in this case, the Earth)
- R is the radius of the celestial body (in this case, the Earth)
3. Effect of Increased Radius and Mass:
If the radius of the Earth is 2 times larger, the new radius (R') would be 2R. Similarly, if the mass of the Earth is 8 times larger, the new mass (M') would be 8M.
4. Calculation of New Escape Velocity:
Using the formula for escape velocity, we can calculate the new escape velocity (Ve'):
Ve' = √(2G(M')/(R'))
Substituting the values of R' and M' in terms of R and M:
Ve' = √(2G(8M)/(2R))
Simplifying the equation:
Ve' = √(8GM/R)
Since Ve' = 2Ve (as per the given condition), we have:
2Ve = √(8GM/R)
Squaring both sides:
4Ve^2 = 8GM/R
Simplifying further:
Ve^2 = 2GM/R
Taking square root on both sides:
Ve = √(2GM/R)
Comparing this with the original formula for escape velocity, we can see that the new escape velocity (Ve') is equal to the square root of 2 times the original escape velocity (Ve).
So, if the radius of the Earth were 2 times larger and the mass 8 times larger, the new escape velocity (Ve') would be 22.4 km/s, which is 2 times the original escape velocity of 11.2 km/s.
Therefore, the correct answer is option 'C' (22.4 km/s).
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If a projectile lifts off from the surface of the Earth with a speed of 11.2 km.s–1, then it can escape from the Earth’s gravitational field completely. This is called the escape velocity. If the radius of the Earth were 2 times larger and the mass 8 times larger, then the escape velocity (in km.s–1) would bea)5.6b)11.2c)22.4d)44.8Correct answer is option 'C'. Can you explain this answer?
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If a projectile lifts off from the surface of the Earth with a speed of 11.2 km.s–1, then it can escape from the Earth’s gravitational field completely. This is called the escape velocity. If the radius of the Earth were 2 times larger and the mass 8 times larger, then the escape velocity (in km.s–1) would bea)5.6b)11.2c)22.4d)44.8Correct answer is option 'C'. Can you explain this answer? for IIT JAM 2024 is part of IIT JAM preparation. The Question and answers have been prepared according to the IIT JAM exam syllabus. Information about If a projectile lifts off from the surface of the Earth with a speed of 11.2 km.s–1, then it can escape from the Earth’s gravitational field completely. This is called the escape velocity. If the radius of the Earth were 2 times larger and the mass 8 times larger, then the escape velocity (in km.s–1) would bea)5.6b)11.2c)22.4d)44.8Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for IIT JAM 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for If a projectile lifts off from the surface of the Earth with a speed of 11.2 km.s–1, then it can escape from the Earth’s gravitational field completely. This is called the escape velocity. If the radius of the Earth were 2 times larger and the mass 8 times larger, then the escape velocity (in km.s–1) would bea)5.6b)11.2c)22.4d)44.8Correct answer is option 'C'. Can you explain this answer?.
If a projectile lifts off from the surface of the Earth with a speed of 11.2 km.s–1, then it can escape from the Earth’s gravitational field completely. This is called the escape velocity. If the radius of the Earth were 2 times larger and the mass 8 times larger, then the escape velocity (in km.s–1) would bea)5.6b)11.2c)22.4d)44.8Correct answer is option 'C'. Can you explain this answer? for IIT JAM 2024 is part of IIT JAM preparation. The Question and answers have been prepared according to the IIT JAM exam syllabus. Information about If a projectile lifts off from the surface of the Earth with a speed of 11.2 km.s–1, then it can escape from the Earth’s gravitational field completely. This is called the escape velocity. If the radius of the Earth were 2 times larger and the mass 8 times larger, then the escape velocity (in km.s–1) would bea)5.6b)11.2c)22.4d)44.8Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for IIT JAM 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for If a projectile lifts off from the surface of the Earth with a speed of 11.2 km.s–1, then it can escape from the Earth’s gravitational field completely. This is called the escape velocity. If the radius of the Earth were 2 times larger and the mass 8 times larger, then the escape velocity (in km.s–1) would bea)5.6b)11.2c)22.4d)44.8Correct answer is option 'C'. Can you explain this answer?.
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