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The ratio of escape velocity at earth (ve) to the escape velocity at a planet (vp) whose radius and mean density are twice as that of earth is
- a)1 : 4
- b)1 : √2
- c)1 : 2
- d)1 : 2√2
Correct answer is option 'D'. Can you explain this answer?
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The ratio of escape velocity at earth (ve) to the escape velocity at ...
As escape velocity.
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The ratio of escape velocity at earth (ve) to the escape velocity at ...
Escape Velocity:
Escape velocity is the minimum velocity needed for an object to escape from the gravitational pull of a celestial body. It is the velocity required to overcome the gravitational attraction and leave the planet's surface without any further propulsion.
Formula for Escape Velocity:
The formula for escape velocity is given by:
ve = √(2GM/R)
Where,
ve = escape velocity
G = gravitational constant
M = mass of the celestial body
R = radius of the celestial body
Ratio of Escape Velocities:
To find the ratio of escape velocities, we need to compare the escape velocity at Earth (ve) with the escape velocity at another planet (vp) with twice the radius and mean density of Earth.
Step 1: Comparing Radii:
Let's assume the radius of Earth as R1 and the radius of the other planet as R2. According to the question, R2 = 2R1.
Step 2: Comparing Mean Densities:
Let's assume the mean density of Earth as ρ1 and the mean density of the other planet as ρ2. According to the question, ρ2 = 2ρ1.
Step 3: Comparing Escape Velocities:
Using the formula for escape velocity, we can write:
ve = √(2GM/R1)
vp = √(2GM/(2R1))
To find the ratio of escape velocities, we can divide the two equations:
ve/vp = (√(2GM/R1)) / (√(2GM/(2R1)))
Simplifying the equation, we get:
ve/vp = √((2GM/R1) / (2GM/(2R1)))
ve/vp = √((2R1) / (2R1))
ve/vp = √1
ve/vp = 1
Therefore, the ratio of escape velocity at Earth (ve) to the escape velocity at a planet (vp) whose radius and mean density are twice as that of Earth is 1:1.
The correct answer is not mentioned in the options provided.
Escape velocity is the minimum velocity needed for an object to escape from the gravitational pull of a celestial body. It is the velocity required to overcome the gravitational attraction and leave the planet's surface without any further propulsion.
Formula for Escape Velocity:
The formula for escape velocity is given by:
ve = √(2GM/R)
Where,
ve = escape velocity
G = gravitational constant
M = mass of the celestial body
R = radius of the celestial body
Ratio of Escape Velocities:
To find the ratio of escape velocities, we need to compare the escape velocity at Earth (ve) with the escape velocity at another planet (vp) with twice the radius and mean density of Earth.
Step 1: Comparing Radii:
Let's assume the radius of Earth as R1 and the radius of the other planet as R2. According to the question, R2 = 2R1.
Step 2: Comparing Mean Densities:
Let's assume the mean density of Earth as ρ1 and the mean density of the other planet as ρ2. According to the question, ρ2 = 2ρ1.
Step 3: Comparing Escape Velocities:
Using the formula for escape velocity, we can write:
ve = √(2GM/R1)
vp = √(2GM/(2R1))
To find the ratio of escape velocities, we can divide the two equations:
ve/vp = (√(2GM/R1)) / (√(2GM/(2R1)))
Simplifying the equation, we get:
ve/vp = √((2GM/R1) / (2GM/(2R1)))
ve/vp = √((2R1) / (2R1))
ve/vp = √1
ve/vp = 1
Therefore, the ratio of escape velocity at Earth (ve) to the escape velocity at a planet (vp) whose radius and mean density are twice as that of Earth is 1:1.
The correct answer is not mentioned in the options provided.
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The ratio of escape velocity at earth (ve) to the escape velocity at a planet (vp) whose radius and mean density are twice as that of earth isa)1 : 4b)1 : √2c)1 : 2d)1 : 2√2Correct answer is option 'D'. Can you explain this answer?
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