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Gravitation Test - 28

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Gravitation Test - 28
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  • Question 1
    1 / -0
    The Earth is assumed to be a sphere of radius R. A platform is arranged at a height R from the surface of the Earth. The escape velocity of a body from this platform is fv, where v is its escape velocity from the surface of the Earth. The value of f is:
    Solution
    According to question and by using COME
    GMmR+R+12m(fv)2-\dfrac{GMm} {R + R} +\dfrac{1} {2}m\left (fv \right )^{2} = 0+00 + 0
    fv\Rightarrow fv = GMR\sqrt{\dfrac{GM} {R}} but v=2GMRv = \sqrt{\dfrac{2GM} {R}}
    Therefore f2GMRf\sqrt{\dfrac{2GM} {R}}GMRf\sqrt{\dfrac{GM} {R}} \Rightarrow f = 12\sqrt{\dfrac{1} {2}}
  • Question 2
    1 / -0
    At a place, the value of 'g' is less by 1% than its value on the surface of the Earth (Radius of Earth, R=6400kmR=6400 km). The place is :
    Solution
    Let acceleration due to gravity at a place be gdg_d. If g is acceleration due to gravity at the surface then,
    gd=g(1dR)g_d=g(1-\frac {d}{R})
    0.99g=g(1dR)0.99g=g(1-\frac {d}{R})
    dR=(0.01)\frac {d}{R}=(0.01)
    d=0.01×6400=64kmd=0.01\times 6400=64 km below the surface of earth
  • Question 3
    1 / -0
    Depth from the surface of the earth at which is acceleration due to gravity is 25% of acceleration due to gravity at the surface
    Solution
    g at the surface is given as gs=43πGRρg_s=\dfrac{4}{3}\pi GR\rho
    and at a depth d it is given as gd=43πG(Rd)ρg_d=\dfrac{4}{3}\pi G(R-d)\rho
    Thus gdgs=RdR\dfrac{g_d}{g_s}=\dfrac{R-d}{R}
    Thus for the depth where acceleration is 25% of the surface gravity we get gdg_d as g4\dfrac{g}{4}
    or
    g4=g(1dR)d=34×R= 4800km\dfrac{g}{4}=g\left ( 1-\dfrac{d}{R} \right )\Rightarrow d=\dfrac{3}{4}\times R = 4800 km    (R is 6371 km)
  • Question 4
    1 / -0
    The variation of g  with height or depth (r) is shown correctly by the graph in the figure (where R = radius of the earth),
    Solution
    ghg_h at height h is given by gh=GM(R+h)2=GM(R+h)2g_h = \dfrac{GM}{(R+h)^2} = GM(R+h)^{-2} which denotes a parabola
    gxg_x at depth x is given by
    gx=g[RxR]g_x = g \left [ \dfrac{R-x}{R} \right ]
    Which denotes is straight line. So, option A is correct.
  • Question 5
    1 / -0
    A man weighs 60kg60 kg at earth's surface. At what height above the earth's surface his weight becomes 30kg30 kg? (radius of earth  =6400km=6400 km).
    Solution
    Above the earth's surface, g1=g(1+hR)2g_1 = \dfrac{g}{(1+\dfrac{h}{R})^2}
    But changed weight = half of its weight on the earth.
    g1=g2\Rightarrow g_1 = \dfrac{g}{2}
    h=2624km.\therefore h = 2624 km.
  • Question 6
    1 / -0
    At what altitude will the acceleration due to gravity be 25%25\% of that at the earth's surface ((given radius of earth is R)R)?
    Solution

    Hint : Use the formula g=GM(R+h)2g'=\dfrac{GM}{(R+h)^{2}}

    Step1: Expression of acceleration due to gravity on earth’s surface\textbf{Step1: Expression of acceleration due to gravity on earth's surface}

    Force on the body placed on Earth's surface is
    F=GMmR2F = \dfrac{GMm}{R^2}
    But, F=mgF = mg hence,
    g= GMmR2g = \dfrac{GMm}{R^2}
    where, variables have their usual meanings.
    Step2: Calculation of altitude\textbf{Step2: Calculation of altitude}
    Now, force on the body at geo-potential  height say hh (altitude) where the acceleration due to gravity is 25%25\% of that at the earth's surface i.e.
    25g100= g4\dfrac{25 g}{100} = \dfrac{g}{4}

    use the formula g=GM(R+h)2g'=\dfrac{GM}{(R+h)^{2}}
    Hence, we can write
    g4=GM(R+h)2\dfrac{g}{4} = \dfrac{GM}{(R+h)^2}

    g4=gR2(R+h)2\dfrac{g}{4} = \dfrac{gR^2}{(R+h)^2}

    (R+h)2=4R2(R+h)^2 = 4R^2
    Taking roots for both sides we get
    R+h=2RR+h = 2R
    h=Rh = R

     

  • Question 7
    1 / -0
    When a satellite moves around the earth in a certain orbit, the quantity which remains constant is:
    Solution

    Hint: Use the Keplers law

    Explanation:

    According to Kepler's rules of planetary motion, when a satellite/planet moves around the Earth/Sun in a certain orbit, aerial velocity remains constant.  

    Correct option is (C) areal velocity

  • Question 8
    1 / -0
    Let ω\omega be the angular velocity of the earth's rotation about its axis. Assume that the acceleration due to gravity on the earth's surface has the same value at the equator and the poles. An object weighed at the equator gives the same reading as a reading taken at a depth a below earth's surface at a pole (d<<R)(d << R) The value of dd is
    Solution
    Let  geg_e be the acceleration due to gravity on surface of earth, gg' be  the acceleration due to gravity at depth dd below the surface of earth and is given by
    g=ge(1dR)g^{'} = g_e \left(1 - \dfrac{d}{R} \right) ............................(1)
    The acceleration due to gravity of an object on latitude(below the surface of earth) is
    g=geω2Rcosϕg^{'} = g_e - \omega^2 R \cos \phi
    where, ϕ\phi is the solid angle at the center of earth sustained by the object.
    At the equator, ϕ=0\phi = 0^\circ
    cosϕ=1\therefore \cos \phi = 1
    Hence, 
    g=geω2Rg' = g_e - \omega^2 R ...............(2)
    From (1) and (2), we can write
     ge(1dR)= geω2R g_e \left(1 - \dfrac{d}{R} \right) = g_e - \omega^2 R 

    dgeR=  ω2R \Rightarrow\dfrac{dg_e}{R} =  \omega^2 R

     d=ω2Rge \Rightarrow d = \dfrac{\omega^2 R}{g_e}

  • Question 9
    1 / -0
    The condition for a uniform spherical mass MM of radius rr to be a black hole is (G=(G= gravitational constant, g=g= acceleration due to gravity )).
    Solution
    For black hole, light even can not escape from the surface of the star.
    Since vesc=2GMrv_{esc} = \sqrt{ \dfrac{2GM}{r}}, putting vesc=cv_{esc} =c where c is the velocity of light c=2GMrc = \sqrt{ \dfrac{2GM}{r}},
    Hence, 2GMrc\sqrt{ \dfrac{2GM}{r}} \ge c
  • Question 10
    1 / -0
    A spring balance is graduated on sea level. If a body is weighted at consecutively increasing heights from earth's surface, the weight indicated by the balance :
    Solution
    A spring balance is graduated on sea level. If a body is weighed with at consecutively increasing heights from earth's surface, the weight indicated by the balance will go on decreasing continuously. This is because value of g decreases with height.
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