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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:√2
- b)1:2
- c)1:2√2
- d)1:4
Correct answer is option 'C'. Can you explain this answer?
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The ratio of escape velocity at earth (ve) to the escape velocity at a...
Calculation of escape velocity for Earth and the planet
Escape velocity is the minimum velocity required by an object to escape from the gravitational pull of a celestial body. It is given by the formula:
VE = (2GM/R)^(1/2)
Where VE is the escape velocity, G is the gravitational constant, M is the mass of the celestial body, and R is the radius of the celestial body.
For Earth, the values of G, M, and R are 6.67 × 10^-11 Nm^2/kg^2, 5.97 × 10^24 kg, and 6.38 × 10^6 m, respectively. Substituting these values in the above formula, we get:
VE = (2 × 6.67 × 10^-11 × 5.97 × 10^24 / 6.38 × 10^6)^(1/2) = 11.2 km/s
For the planet whose radius and mean density are twice that of Earth, the values of G, M, and R are the same as that of Earth, but the mass is twice and the radius is twice. Substituting these values in the above formula, we get:
VP = (2 × 6.67 × 10^-11 × 2 × 5.97 × 10^24 / 2 × 2 × 6.38 × 10^6)^(1/2) = 7.9 km/s
Calculation of the ratio of escape velocities
The ratio of escape velocities is given by:
VE/VP = 11.2/7.9 = 1.41 ≈ 1.4
Therefore, the correct option is C) 1:2.
Escape velocity is the minimum velocity required by an object to escape from the gravitational pull of a celestial body. It is given by the formula:
VE = (2GM/R)^(1/2)
Where VE is the escape velocity, G is the gravitational constant, M is the mass of the celestial body, and R is the radius of the celestial body.
For Earth, the values of G, M, and R are 6.67 × 10^-11 Nm^2/kg^2, 5.97 × 10^24 kg, and 6.38 × 10^6 m, respectively. Substituting these values in the above formula, we get:
VE = (2 × 6.67 × 10^-11 × 5.97 × 10^24 / 6.38 × 10^6)^(1/2) = 11.2 km/s
For the planet whose radius and mean density are twice that of Earth, the values of G, M, and R are the same as that of Earth, but the mass is twice and the radius is twice. Substituting these values in the above formula, we get:
VP = (2 × 6.67 × 10^-11 × 2 × 5.97 × 10^24 / 2 × 2 × 6.38 × 10^6)^(1/2) = 7.9 km/s
Calculation of the ratio of escape velocities
The ratio of escape velocities is given by:
VE/VP = 11.2/7.9 = 1.41 ≈ 1.4
Therefore, the correct option is C) 1:2.
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The ratio of escape velocity at earth (ve) to the escape velocity at a planet (vp) whose radius and meandensity are twice as that of earth is :a)1:√2b)1:2c)1:2√2d)1:4Correct answer is option 'C'. Can you explain this answer?
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