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    Gravitation: 4 Problems

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    1) The following formula represents the period of a pendulum,T.
    T = 2π(l/g)
    (a) What would be the period of a 1.8 m long pendulum on the moon's surface? The moon's mass is 7.34 1022 kg, and its radius is 1.74x10^6 m.

    2) A force of 40.9 N is required to pull a 10.2 kg wooden block at a constant velocity across a smooth glass surface on Earth. What force would be required to pull the same wooden block across the same glass surface on the planet Jupiter?

    3) A 1.1 kg mass weighs 10.78 N on Earth's surface, and the radius of Earth is roughly 6.4 106 m. (Use G = 6.67 10-11 N·m2/kg2.)
    (a) Calculate the mass of Earth.
    kg
    (b) Calculate the average density of Earth.
    kg/m3

    4) An apparatus like the one Cavendish used to find G has a large lead ball that is 6.2 kg in mass and a small one that is 0.040 kg. Their centers are separated by 0.055 m. Find the force of attraction between them. (Use G = 6.67 10-11 N·m2/kg2.)
    N

    © BrainMass Inc. brainmass.com December 24, 2021, 8:28 pm ad1c9bdddf
    https://brainmass.com/physics/gravity/gravitation-problems-279564

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    1) The following formula represents the period of a pendulum.
    T = 2π(l/g)
    (a) What would be the period of a 1.8 m long pendulum on the moon's surface? The moon's mass is 7.34 1022 kg, and its radius is 1.74x10^6 m.
    The time period T of a simple pendulum of length l is given by

    Here g is the acceleration due to gravity at the place of experiment.
    Now the acceleration due to gravity at the surface of a planet is given by

    Here G is universal gravitation constant, M is the mass of the planet and R is the radius of the planet.
    Thus the acceleration due to gravity near the surface of moon will be given by the same formula.
    Here mass of the moon M = 7.34*1022 kg
    Radius of moon R = 1.74*106 m
    And gravitation constant G = 6.67*10-11 N m2/kg2

    Thus the acceleration at the surface of moon will be
    m/s2
    Thus the time period of the pendulum of length l = 1.80 m on the surface of moon is given by

    Thus the period of small oscillation of pendulum will be 6.6 s.

    2) A force of 40.9 N is required to pull a 10.2 kg wooden block at a constant velocity across a smooth glass surface on Earth. What force would be required to pull the same wooden block across the same glass surface on the planet Jupiter?
    The maximum force of kinetic friction between two surfaces is the given by

    Here is the coefficient of kinetic friction and N is the normal reaction between the surfaces. With the block on the horizontal surface it is equal to the weight mg of the block.
    As the block is moving on the surface with constant velocity (zero acceleration), the force applied F is just equal to the kinetic friction and thus we have

    Or ----------------------- (1)
    Now the acceleration due to gravity g' at the surface of Jupiter will not be same as that of earth and thus the required force at the surface of Jupiter is given by
    ----------------------- (2)
    Dividing equation (2) by equation (1) we get

    Or ----------------------- (3)
    Now the acceleration due to gravity at the surface of Jupiter is given by

    Here G is universal gravitation constant, M' is the mass of Jupiter and R' is its radius.
    Thus the acceleration due to gravity at the surface of Jupiter will be given by the same formula.
    Here mass of the moon M' = 1.90*1027 kg
    Radius of moon R' = 7.15*107 m
    And gravitation constant G = 6.67*10-11 N m2/kg2

    Thus the acceleration at the surface Jupiter will be
    m/s2
    Substituting the value of g = 9.81 m/s2 at the surface of earth and g' = 24.8 m/s2 at the surface of Jupiter in equation (3) we get

    Thus the force requited the same block at the surface of Jupiter will be 103 N.

    3) A 1.1 kg mass weighs 10.78 N on Earth's surface, and the radius of Earth is roughly 6.4 106 m. (Use G = 6.67 10-11 N·m2/kg2.)
    (a) Calculate the mass of Earth.
    kg
    The weight of a body is the force of gravity acting on the body and is given by
    W = mg
    Thus the acceleration due to gravity at the surface of the earth is given by
    g = W/m = 10.78/1.1 = 9.8 m/s2
    Now as the acceleration due to gravity at the surface of earth is given by

    Substituting the given values we get

    Gives
    Or
    Thus the mass of earth will be 6.02*1024 kg.

    (b) Calculate the average density of Earth.
    kg/m3
    The density is the mass per unit volume of a substance and thus the average density of earth is given by the mass of the earth divided by volume of the earth.

    Substituting numerical values se get
    kg/m3
    Thus the density of earth will be 5.5*103 kg/m3

    4) An apparatus like the one Cavendish used to find G has a large lead ball that is 6.2 kg in mass and a small one that is 0.040 kg. Their centers are separated by 0.055 m. Find the force of attraction between them. (Use G = 6.67 10-11 N·m2/kg2.)
    N

    According to Newton's law of gravitation the magnitude of gravitational force of attraction between the two point-masses of mass m1 and m2 at a distance r is given by

    As a spherical body can be considered as point mass in terms of gravitational outside it the force between two spheres is given by substituting the values as

    Thus the force between the two balls will be 5.47*10-9 N.

    This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!

    © BrainMass Inc. brainmass.com December 24, 2021, 8:28 pm ad1c9bdddf>
    https://brainmass.com/physics/gravity/gravitation-problems-279564

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