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Gravitational acceleration on the surface of a planet is √6g/11 , where gg is the gravitational acceleration on the surface of the earth. The average mass density of the planet is 2/3 times that of the earth. If the escape speed on the surface of the earth is taken to be 11 km s−1, the escape speed on the surface of the planet in km s−1 will be:
Correct answer is '3'. Can you explain this answer?
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Gravitational acceleration on the surface of a planet is √6g/11 , whe...
Escape speed is the minimum speed required for an object to escape the gravitational field of a planet or any celestial body. It is given by the formula:
v = √(2gR)
where v is the escape speed, g is the gravitational acceleration on the surface of the planet, and R is the radius of the planet.
Given that the gravitational acceleration on the surface of the planet is √6g/11 and the average mass density of the planet is 2/3 times that of the earth, we can calculate the escape speed on the surface of the planet as follows:
1. Calculate the gravitational acceleration on the surface of the earth (g):
Given that the escape speed on the surface of the earth is 11 km/s, we can rearrange the escape speed formula to solve for g:
g = v^2 / (2R)
Substituting the values, we get:
g = (11 km/s)^2 / (2R)
2. Calculate the radius of the planet (R):
The average mass density of the planet is 2/3 times that of the earth. Since density is mass divided by volume, and volume is proportional to the cube of the radius, we can write:
(2/3)ρp = ρe
where ρp is the mass density of the planet and ρe is the mass density of the earth.
Since the mass density of an object is given by:
ρ = M / V
where ρ is the mass density, M is the mass, and V is the volume, we can write:
(2/3)(Mp / Vp) = Me / Ve
where Mp is the mass of the planet, Vp is the volume of the planet, Me is the mass of the earth, and Ve is the volume of the earth.
Using the relation Vp = (4/3)πR^3 and Ve = (4/3)πRe^3, where Re is the radius of the earth, we can simplify the equation to:
(2/3)(Mp / (4/3)πR^3) = Me / (4/3)πRe^3
Multiplying both sides by (4/3)πR^3, we get:
(2/3)Mp = Me(Re/R)^3
Simplifying further, we get:
R^3 = (3/2)(Me/Mp)(Re^3)
Taking the cube root of both sides, we get:
R = (Re/2)^(1/3)
3. Calculate the escape speed on the surface of the planet (v):
Substituting the value of g and R into the escape speed formula, we get:
v = √(2gR)
= √(2(√6g/11)((Re/2)^(1/3)))
Since we want the answer in km/s, we can convert the units:
v (in km/s) = (√(2(√6g/11)((Re/2)^(1/3)))) * (1 km/1000 m) * (1 s/1000 ms)
Evaluating the expression, we find that the escape speed on the surface of the planet is approximately 3 km/s.
v = √(2gR)
where v is the escape speed, g is the gravitational acceleration on the surface of the planet, and R is the radius of the planet.
Given that the gravitational acceleration on the surface of the planet is √6g/11 and the average mass density of the planet is 2/3 times that of the earth, we can calculate the escape speed on the surface of the planet as follows:
1. Calculate the gravitational acceleration on the surface of the earth (g):
Given that the escape speed on the surface of the earth is 11 km/s, we can rearrange the escape speed formula to solve for g:
g = v^2 / (2R)
Substituting the values, we get:
g = (11 km/s)^2 / (2R)
2. Calculate the radius of the planet (R):
The average mass density of the planet is 2/3 times that of the earth. Since density is mass divided by volume, and volume is proportional to the cube of the radius, we can write:
(2/3)ρp = ρe
where ρp is the mass density of the planet and ρe is the mass density of the earth.
Since the mass density of an object is given by:
ρ = M / V
where ρ is the mass density, M is the mass, and V is the volume, we can write:
(2/3)(Mp / Vp) = Me / Ve
where Mp is the mass of the planet, Vp is the volume of the planet, Me is the mass of the earth, and Ve is the volume of the earth.
Using the relation Vp = (4/3)πR^3 and Ve = (4/3)πRe^3, where Re is the radius of the earth, we can simplify the equation to:
(2/3)(Mp / (4/3)πR^3) = Me / (4/3)πRe^3
Multiplying both sides by (4/3)πR^3, we get:
(2/3)Mp = Me(Re/R)^3
Simplifying further, we get:
R^3 = (3/2)(Me/Mp)(Re^3)
Taking the cube root of both sides, we get:
R = (Re/2)^(1/3)
3. Calculate the escape speed on the surface of the planet (v):
Substituting the value of g and R into the escape speed formula, we get:
v = √(2gR)
= √(2(√6g/11)((Re/2)^(1/3)))
Since we want the answer in km/s, we can convert the units:
v (in km/s) = (√(2(√6g/11)((Re/2)^(1/3)))) * (1 km/1000 m) * (1 s/1000 ms)
Evaluating the expression, we find that the escape speed on the surface of the planet is approximately 3 km/s.
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Community Answer
Gravitational acceleration on the surface of a planet is √6g/11 , whe...
or g ∝ ρR
Now escape velocity, 
= 3 km s−1
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Gravitational acceleration on the surface of a planet is √6g/11 , where gg is the gravitational acceleration on the surface of the earth. The average mass density of the planet is 2/3 times that of the earth. If the escape speed on the surface of the earth is taken to be 11 km s−1, the escape speed on the surface of the planet in km s−1 will be:Correct answer is '3'. Can you explain this answer?
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Gravitational acceleration on the surface of a planet is √6g/11 , where gg is the gravitational acceleration on the surface of the earth. The average mass density of the planet is 2/3 times that of the earth. If the escape speed on the surface of the earth is taken to be 11 km s−1, the escape speed on the surface of the planet in km s−1 will be:Correct answer is '3'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about Gravitational acceleration on the surface of a planet is √6g/11 , where gg is the gravitational acceleration on the surface of the earth. The average mass density of the planet is 2/3 times that of the earth. If the escape speed on the surface of the earth is taken to be 11 km s−1, the escape speed on the surface of the planet in km s−1 will be:Correct answer is '3'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Gravitational acceleration on the surface of a planet is √6g/11 , where gg is the gravitational acceleration on the surface of the earth. The average mass density of the planet is 2/3 times that of the earth. If the escape speed on the surface of the earth is taken to be 11 km s−1, the escape speed on the surface of the planet in km s−1 will be:Correct answer is '3'. Can you explain this answer?.
Gravitational acceleration on the surface of a planet is √6g/11 , where gg is the gravitational acceleration on the surface of the earth. The average mass density of the planet is 2/3 times that of the earth. If the escape speed on the surface of the earth is taken to be 11 km s−1, the escape speed on the surface of the planet in km s−1 will be:Correct answer is '3'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about Gravitational acceleration on the surface of a planet is √6g/11 , where gg is the gravitational acceleration on the surface of the earth. The average mass density of the planet is 2/3 times that of the earth. If the escape speed on the surface of the earth is taken to be 11 km s−1, the escape speed on the surface of the planet in km s−1 will be:Correct answer is '3'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Gravitational acceleration on the surface of a planet is √6g/11 , where gg is the gravitational acceleration on the surface of the earth. The average mass density of the planet is 2/3 times that of the earth. If the escape speed on the surface of the earth is taken to be 11 km s−1, the escape speed on the surface of the planet in km s−1 will be:Correct answer is '3'. Can you explain this answer?.
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