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Gravitation Test - 7

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Gravitation Test - 7
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  • Question 1
    1 / -0
    Gravity is a conservative force. Which of the following is incorrect about this force?
    Solution
    Kinetic energy depends upon the speed and mass of the body. More massive objects would have more kinetic energy than other objects of the same mass.
  • Question 2
    1 / -0
    The dimensional formula for gravitational constant is
    Solution
    Newton`s law of gravitation F =
    So, G =
    where, F is the force between the two masses m1, m2 at a distance r
    Dimensional formula of force = M1 L1 T-2
    Dimensional formula of gravitational constant = M-1L3 T-2
  • Question 3
    1 / -0
    Ball pens function on the principle of
    Solution
    Ball pens function on the principle of gravity. Fountain pens function on the principle of capillarity and gravity.
  • Question 4
    1 / -0
    At what height above the earth's surface does the acceleration due to gravity fall to 1% of its value at the earth's surface?
    Solution
    = 9.8 m/s2
    = 1.1% of g means the denominator of the first equation will be 100 times R2

    100R2 = (10R)2 = (h + R)2 (where h is the altitude)

    10R = h + R

    9R = h
  • Question 5
    1 / -0
    As we go from the equator to the poles, the value of g
    Solution
    The value of 'g' is minimum at the equator and maximum at the poles.
    It means 'g' increases as we go from the equator towards the poles.
    This is due to the diference in the the radius of the earth, which is greater at the equator than at the poles due to slight bulge in the earth at the equator and slight flattening of it at the poles.
  • Question 6
    1 / -0
    The value of g at a particular point is 9.8 m/s2. Suppose, the Earth suddenly shrinks uniformly to half its present size without losing any mass. The value of 'g' at the same point (assuming that the distance of the point from the centre of the Earth does not shrink) will now be
    Solution
    Consider a point A outside the Earth where value of g = 9.8 m/s2, as shown in the figure



    Let height of the point from the surface of the Earth be h.
    Let radius of the Earth be R.
    Acceleration due to gravity at height h from the surface of the Earth is
    g` = (M is the mass)
    Consider the height of the point when the Earth has shrunken be .
    New radius will be , as shown in the figure.



    Acceleration due to gravity at this point will be
    g" =




    = 9.8 m/s2
  • Question 7
    1 / -0
    The acceleration due to earth`s gravity ______ when we go a little distance inside the earth. It ______ when we go a little distance above the earth.
    Solution
    The value of g decreases on going below the surface of the earth.
    It is because (where d is the depth)
    And, as we go above the surface of the earth, the value of g decreases because,
  • Question 8
    1 / -0
    An object is at a height R/2 above the Earth. What is the value of acceleration due to gravity at that instant?
    Solution
    Consider an object at a height from the Earth's surface, where R = radius of the Earth.
    Value of acceleration due to gravity at that instant will be:
    g' = ; where M = mass of the Earth

    Or g' = 4/9(GM/R2)
    Or g' = (4/9)g
    Hence, option 2 is correct.
  • Question 9
    1 / -0
    The path of an object released from earth at a velocity less than escape velocity and more than orbital velocity is called
    Solution
    The path of an object released from earth at a velocity less than escape velocity and more than orbital velocity is called incomplete Elliptical.
  • Question 10
    1 / -0
    Escape velocity from earth is approximately
    Solution
    It is the least velocity required to throw a body away from the surface of the earth so that it may not return.
    Escape velocity from the earth is approximately 11.2 km/s.
  • Question 11
    1 / -0
    The escape velocity for a planet is ve. A particle starts from rest at large distance from the planet and reaches the planet only under gravitational attraction. And then it passes through a smooth tunnel through its centre. Its speed at the centre of the planet will be
    Solution
    Taking the potential at a large distance from the planet as zero, the potential at the centre of the planet =

    V2 =
    V =
  • Question 12
    1 / -0
    The time period of a pendulum is independent of the
    Solution
    Time period of a pendulum is given by:
    , where L is the pendulum length and
    g is the acceleration due to gravity.
    So, the mass of the bob does not affect the time period.
  • Question 13
    1 / -0
    Let vo be the velocity of orbit around a planet. The escape velocity of the planet is
    Solution
    Escape velocity is times the orbital velocity.
    Vescape = Vorbit
  • Question 14
    1 / -0
    What is the orbital velocity of a satellite orbiting the earth just at the surface?
    Solution
    Vescape = Vorbital
    We know that, Vescape = 11.2 km/s
    11.2 = Vorbital
    = Vorbital
    Vorbital = 7.94 km/s
    7.9 km/s
  • Question 15
    1 / -0
    If a small part separates from an orbiting satellite, the part will
    Solution
    The small part will continue to move in the same orbit after separating from its satellite because of the orbital velocity. The part achieves balance between gravity's pull on it and the inertia of motion resulting in its sustained motion.
  • Question 16
    1 / -0
    The path of a planet around the sun is elliptical as shown in the figure. Its velocity is greatest at

    Solution
    In an elliptical orbit, the velocity is greatest at the perigee of the orbit. It means, when the object is closest to the Sun.
  • Question 17
    1 / -0
    Compare the area swept by the line joining planet and sun at apogee (closest point in orbit) and perigee (farthest point in orbit) during a time interval `T`.
    Solution
    Kepler`s second law states that the area swept by all lines joining a satellite and the Earth is equal, during a given interval of time. So, the rate f, at which the satellite sweeps, is constant. The same will be followed in the question.
  • Question 18
    1 / -0
    The radius of a planet's orbit increases by a factor of three. Correspondingly, time period becomes
    Solution
    T2 is directly proportional to R3.
    The radius of a planet's orbit increases by a factor of three.
    So, T2 = K(4R)3
    If K = 1, then
    T = 8R3/2
  • Question 19
    1 / -0
    The ratio of the inertial to gravitational mass is equal to
    Solution
    The ratio of the interval and gravitational mass is equal to one. The equivalence of inertial and gravitational masses is also referred to as `Galilean Equivalence Principle`.
  • Question 20
    1 / -0
    Which of the following statements is correct?
    Solution
    The weight of the body is taken to be the force on the body due to gravity whereas, mass is the measure of inertia or a fundamental measure of the amount of matter in the body.
    Therefore, the weight of a body can be changed without changing its mass.
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