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A man can jump vertically to a height of 1.5m  on the earth. Calculate the radius of a planet (in kms ) of the same mean density as that of the earth from whose gravitational field he could escape by jumping. Radius of earth is 6.41 × 106m.
    Correct answer is '3.1'. Can you explain this answer?
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    A man can jump vertically to a height of 1.5mon the earth. Calculate t...
    For the asked planet this u should be equal to the escape velocity from its surface.

    The correct answer is: 3.1
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    A man can jump vertically to a height of 1.5mon the earth. Calculate t...
    To calculate the radius of another planet from which the man could escape by jumping, we need to equate the gravitational potential energy on the surface of the planet (where he jumps) to the kinetic energy he gains while jumping.

    The gravitational potential energy is given by the formula:
    PE = mgh

    Where:
    m = mass of the man
    g = acceleration due to gravity on the surface of the planet
    h = height the man jumps

    The kinetic energy gained while jumping is given by the formula:
    KE = (1/2)mv^2

    Where:
    m = mass of the man
    v = velocity gained while jumping

    Since the man jumps vertically, the initial velocity is zero, and the final velocity is given by:
    v = √(2gh)

    Now, let's equate the gravitational potential energy to the kinetic energy:

    mgh = (1/2)mv^2

    Canceling out the mass (m) from both sides, we get:

    gh = (1/2)v^2

    Now, let's substitute the expression for v:

    gh = (1/2)(√(2gh))^2
    gh = (1/2)(2gh)

    Simplifying further:

    gh = gh

    This equation tells us that the height (h) to which the man can jump is the same on both planets, as it only depends on the acceleration due to gravity (g) and the height (h) itself.

    So, we can write:

    1.5m(9.81m/s^2) = h(√(GMe/Re^2))
    1.5(9.81) = h(√(GM/Re^2))

    Where:
    G = gravitational constant
    Me = mass of the Earth
    Re = radius of the Earth

    Now, let's rearrange the equation to solve for Re:

    Re^2 = GM/(1.5(9.81))^2

    Since the density is the same, we can write:

    Me/Re^3 = M/Re^3

    Where:
    M = mass of the other planet

    Thus:

    Re^3 = Me/M * R^3

    Re = (Me/M)^(1/3) * R

    Now, let's substitute the known values:

    Re = (5.972 × 10^24 kg / M)^(1/3) * 6.41 × 10^6 m

    To escape the gravitational field by jumping, the man would need to jump to a height greater than the radius of the planet. Therefore, we can equate Re to h:

    h = (5.972 × 10^24 kg / M)^(1/3) * 6.41 × 10^6 m

    Given that h = 1.5 m, we can solve for M:

    1.5 = (5.972 × 10^24 kg / M)^(1/3) * 6.41 × 10^6

    Cubing both sides:

    1.5^3 = (5.972 × 10^24 kg / M) * (6.41 × 10^6)^3

    Simplifying:

    3.375 = (5.972 × 10^24 kg / M) * (2.607281 × 10^19 kg)

    Now, let's solve for M:

    M = (5
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    A man can jump vertically to a height of 1.5mon the earth. Calculate the radius of a planet (inkms) of the same mean density as that of the earth from whose gravitational field he could escape by jumping. Radius of earth is 6.41 × 106m.Correct answer is '3.1'. Can you explain this answer?
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    A man can jump vertically to a height of 1.5mon the earth. Calculate the radius of a planet (inkms) of the same mean density as that of the earth from whose gravitational field he could escape by jumping. Radius of earth is 6.41 × 106m.Correct answer is '3.1'. Can you explain this answer? for Physics 2024 is part of Physics preparation. The Question and answers have been prepared according to the Physics exam syllabus. Information about A man can jump vertically to a height of 1.5mon the earth. Calculate the radius of a planet (inkms) of the same mean density as that of the earth from whose gravitational field he could escape by jumping. Radius of earth is 6.41 × 106m.Correct answer is '3.1'. Can you explain this answer? covers all topics & solutions for Physics 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A man can jump vertically to a height of 1.5mon the earth. Calculate the radius of a planet (inkms) of the same mean density as that of the earth from whose gravitational field he could escape by jumping. Radius of earth is 6.41 × 106m.Correct answer is '3.1'. Can you explain this answer?.
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