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Gravitation Test - 13

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Gravitation Test - 13
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  • Question 1
    1 / -0

    If there were a smaller gravitational effect, which of the following forces do you think would alter in some respect?

    Solution

    Archimedes uplift as it depends on the weight of the body.

  • Question 2
    1 / -0

    A satellite of the earth is revolving in a circular orbit with a uniform speed v. If the gravitational force suddenly disappears, the satellite will

    Solution

    Due to inertia of direction.

  • Question 3
    1 / -0

    If the distance between two masses is doubled, the gravitational attraction between them

    Solution

    \(F \propto \frac{1}{r^2}\)

    If r becomes double then F reduces to \(\frac{F}{4}\) 

  • Question 4
    1 / -0

    The centripetal force acting on a satellite orbiting round the earth and the gravitational force of earth acting on the satellite both equal F. The net force on the satellite is

    Solution

    Actually gravitational force provides the centripetal force.

  • Question 5
    1 / -0

    The time period of a simple pendulum on a freely moving artificial satellite is

    Solution

    Time period of simple pendulum

    T = \(2\pi \sqrt{\frac{l}{g'}}\)

    In artificial satellite g′ = 0 and 

    \(\therefore\,T\) = infinite.

  • Question 6
    1 / -0

    A body of mass m is taken to the bottom of a deep mine. Then

    Solution

    Because acceleration due to gravity decreases

  • Question 7
    1 / -0

    In order to find time, the astronaut orbiting in an earth satellite should use

    Solution

    In pendulum clock the time period depends on the value of g, while in spring watch, the time period is independent of the value of g.

  • Question 8
    1 / -0

    The weight of an object in the coal mine, sea level, at the top of the mountain is \(W_1, W_2\) and \(W_3\) respectively, then

    Solution

    Because value of g decreases when we move either in coal mine or at the top of mountain.

  • Question 9
    1 / -0

    The mass of the earth is 81 times that of the moon and the radius of the earth is 3.5 times that of the moon. The ratio of the acceleration due to gravity at the surface of the moon to that at the surface of the earth is

    Solution

    g = \(\frac{GM}{R^2}\)

    Given \(M_e\) = 81 \(M_m\)\(R_e\) = 3.5 \(R_m\)) Substituting the above values, \(\frac{g_m}{g_e}\) = 0.15

  • Question 10
    1 / -0

    In a gravitational field, at a point where the gravitational potential is zero

    Solution

    I = \(-\frac{dV}{dx} \)

    If V = 0 then gravitational field is necessarily zero.

  • Question 11
    1 / -0

    There are two bodies of masses 100 kg and 10000 kg separated by a distance 1m. At what distance from the smaller body, the intensity of gravitational field will be zero

    Solution

    \(\frac{G \times 100}{x^2}\) = \(\frac{G \times 10000}{(1 - x)^2}\)

    ⇒ \(\frac{10}{x} = \frac{100}{1 - x}\)

    ⇒ x = \(\frac{1}{11}m\)

  • Question 12
    1 / -0

    Radius of orbit of satellite of earth is R. Its kinetic energy is proportional

    Solution

    K.E. = \(\frac{GMm}{2R}\)

  • Question 13
    1 / -0

    Select the correct statement from the following :

    Solution

    \(v_0 = \sqrt{\frac{GM}{r}}\)

  • Question 14
    1 / -0

    Consider a satellite going round the earth in an orbit. Which of the following statements is wrong?

    Solution

    Centripetal acceleration works on it.

  • Question 15
    1 / -0

    Assertion: Gravitational force between two particles is negligibly small compared to the electrical force.

    Reason: The electrical force is experienced by charged particles only.

    (A) If both assertion and reason are true and the reason is the correct explanation of the assertion.

    (B) If both assertion and reason are true but reason is not the correct explanation of the assertion.

    (C) If assertion is true but reason is false.

    (D) If the assertion and reason both are false.

    Solution

    For two-electron \(\frac{F_g}{F_e} = 10^{-43}\) i.e. gravitational force is negligible in comparison to electrostatic force of attraction.

  • Question 16
    1 / -0

    If the earth is at one-fourth of its present distance from the sun, the duration of the year will be

    Solution

    Since \(T^2 \propto r^3\)

    \(\therefore \Big(\frac{T'}{T} \Big)^2 = \Big(\frac{1}{4} \Big)^3 \)

    ⇒ T' = \(\frac{1}{8}T\)

  • Question 17
    1 / -0

    The earth revolves about the sun in an elliptical orbit with mean radius \(9.3 ×10^7\) m in a period of 1 year. Assuming that there are no outside influences

    Solution

    Kinetic and potential energies vary with position of earth w.r.t. sun. Angular momentum remains constant everywhere.

  • Question 18
    1 / -0

    As observed from the earth, the sun appears to move in an approximately circular orbit. For the motion of another planet like mercury as observed from the earth, this would

    Solution

    As observed from the earth, the sun appears to move in an approximate circular orbit. The gravitational force of attraction between the earth and the sun always follows inverse square law. Due to relative motion between the earth and mercury, the orbit of mercury, as observed from the earth, will not be approximately circular, since the major gravitational force on mercury is due to the sun.

  • Question 19
    1 / -0

    A missile is launched with a velocity less than the escape velocity. The sum of its kinetic and potential energy is

    Solution

    At earth, total energy of the missile is \(\frac{1}{2}mv^2 - \frac{GmR}{r}\)

    Where, mm is the mass of the missile, vv is the velocity of the missile at a distance rr from the earth, RR is the radius of earth. Now to escape the gravitational pull of earth, this total energy should be greater than zero or equal to zero (energy at infinity), i.e.

    \(\frac{1}{2}mv^2 - \frac{GmR}{r} \geq 0\)

  • Question 20
    1 / -0

    For the moon to cease to remain the earth's satellite, its orbital velocity has to increase by a factor of

    Solution

    \(V_e = \sqrt 2 v_0\)

    i.e. if the orbital velocity of moon is increased by factor of \(\sqrt 2\) then it will escape out from the gravitational field of earth.

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