Self Studies
Selfstudy
Selfstudy

Gravitation Test - 76

Result Self Studies

Gravitation Test - 76
  • Score

    -

    out of -
  • Rank

    -

    out of -
TIME Taken - -
Self Studies

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Self Studies Self Studies
Weekly Quiz Competition
  • Question 1
    1 / -0
    If g is the acceleration due to gravity on the surface of earth, find the gain in potential energy of an object of mass m raised from the surface of earth to a height equal to the radius R of the earth.
    Solution
    Let M, R be the mass and radius of the earth. Then
    $$g=GM/R^2$$
    or $$GM=gR^2$$                (i)
    Potential energy of the object on the surface of earth
    $$U_1=\dfrac{-GMm}{R}$$
    The potential energy of the object at a height equal to radius of the earth is
    $$U_2=\dfrac{-GMm}{2R}$$
    Gain in PE is
    $$U_2-U_1=-\dfrac{GMm}{2R}+\dfrac{GMm}{R}=\dfrac{GMm}{2R}$$
    $$=\dfrac{(gR^2)m}{2R}=\dfrac{1}{2}mg R$$.
  • Question 2
    1 / -0
    A skylab of mass m kg is first launched from the surface of the earth in a circular orbit of radius $$2$$R(from the centre of the earth) and then it is shifted from this circular orbit to another circular orbit of radius $$3$$R. The minimum energy required to place the lab in the first orbit and to shift the lab from first orbit to the second orbit are.
    Solution
    Energy required to place the lab in first orbit is the change in potential energy$$+{ k }_{ 0 }{ E }_{ 0 }$$
    $$\triangle PE=\cfrac { -GMm }{ 2R } +\cfrac { GMm }{ 2R } =\cfrac { GMm }{ 2R } =\triangle u$$
    Velocity given to lab is provided by gravitation force. So, centripetal force=gravitational force
    $$\cfrac { m{ v }^{ 2 } }{ 2R } =\cfrac { GMm }{ { \left( 2R \right)  }^{ 2 } } \Rightarrow v=\sqrt { \cfrac { GM }{ 2R }  } $$
    $$KE=\cfrac { 1 }{ 2 } m{ v }^{ 2 }=\cfrac { 1 }{ 2 } \cfrac { GM }{ 2R } =\cfrac { 1 }{ 4 } \cfrac { GMm }{ R } $$
    Total energy$$=\triangle u+{ K.E }_{ 0 }$$
    $$=\cfrac { GMm }{ 2R } +\cfrac { 1 }{ 4 } \cfrac { GMm }{ R } $$
    $$=\cfrac { 3 }{ 4 } \cfrac { GMm }{ R } $$$$g=\cfrac { GMm }{ { R }^{ 2 } } $$
    $$T.E.=\cfrac { 3 }{ 4 } mgR$$
    Similarly to shift to orbit of $$3R$$
    $$\cfrac { m{ v }^{ 2 } }{ 3R } =\cfrac { GMm }{ { \left( 3R \right)  }^{ 2 } } \Rightarrow { v }^{ 2 }=\cfrac { GM }{ 3R } $$
    Total energy$$=\cfrac { GMm }{ R } -\cfrac { GMm }{ 3R } +\cfrac { 1 }{ 2 } m{ v }^{ 2 }$$
    $$=\cfrac { GMm }{ R } -\cfrac { GMm }{ 3R } +\cfrac { 1 }{ 2 } \cfrac { GMm }{ 3R } $$
    $$=\cfrac { GMm }{ R } \left( 1-\cfrac { 1 }{ 3 } +\cfrac { 1 }{ 6 }  \right) $$
    $$=\cfrac { 5 }{ 6 } mgR$$
    Energy required$$=\cfrac { 5 }{ 6 } mgR-\cfrac { 3 }{ 4 } mgR$$
    $$=\cfrac { \left( 20-18 \right) mgR }{ 24 } $$
    Minimum energy required$$=\cfrac { mgR }{ 12 } $$
  • Question 3
    1 / -0
    A point mass m is released from rest at a distance of $$3$$R from the centre of a thin-walled hollow sphere of radius R and mass M as shown. The hollow sphere is fixed in position and the only force on the point mass is the gravitational attraction of the hollow sphere. There is a very small hole in the hollow sphere through which the point mass falls as shown. The velocity of a point mass when it passes through point P at a distance $$R/2$$ from the centre of the sphere is?

    Solution
    Gravitational potential inside hollow sphere $$=-\cfrac{GM}{R}$$
    Gravitational potential at distance $$3R=-\cfrac{GM}{3R}$$
    $$\therefore$$ Conservation of energy
    $$\Rightarrow\cfrac{1}{2}mv^2-\cfrac{GMm}{R}=-\cfrac{GMm}{3R}\\ \Rightarrow {1}{2}mv^2=\cfrac{GMm}{R}(1-\cfrac{1}{3})\\ \Rightarrow\cfrac{v^2}{2}=\cfrac{2GM}{3R}\\ \Rightarrow v^2=\cfrac{4GM}{3R}\\ v=\sqrt{\cfrac{4GM}{3R}}$$

  • Question 4
    1 / -0
    What is the fractional decrease in the value of free-fall acceleration g for a particle when it is lifted from the surface to an elevation h? (h $$< < $$ R).
    Solution
    Given that 
    $$g=\dfrac { GM }{ { R }^{ 2 } } $$
    $$\dfrac { dg }{ dR } =\dfrac { -2GM }{ { R }^{ 3 } } $$
    $$\dfrac { dg }{ h } =\dfrac { -2GM }{ { R }^{ 2 } } .\dfrac { 1 }{ R } $$
    $$\dfrac { dg }{ g } =-2\left( \dfrac { h }{ R }  \right) $$
  • Question 5
    1 / -0
    The density of a newly discovered planet is twice that of the earth. The acceleration due to gravity at the surface of the planet is equal to that at the surface of the earth. If the radius of the earth is R and the radius of the planet is R'. Then what is the value of R/R'?
    Solution
    let,$$\\ { g }^{ \prime  }$$= aceleration due to gravity at newly discovered planet.
    g=acceleration due to gravity at earth
    $$\\ \therefore { g }^{ \prime  }=g\\ =>\cfrac { G{ M }^{ \prime  } }{ { R }^{ \prime 2 } } =\cfrac { GM }{ { R }^{ 2 } } \\ =>\cfrac { G.\cfrac { 4 }{ 3 } \pi { R }^{ \prime 3 }{ p }^{ 1 } }{ { R }^{ \prime 2 } } =\cfrac { G.\cfrac { 4 }{ 3 } \pi { R }^{ 2 }{ p } }{ { R }^{ 2 } } \\ =>{ R }^{ \prime  }{ p }^{ \prime  }=R.p\\ =>\cfrac { R }{ { R }^{ \prime  } } =\cfrac { { p }^{ \prime  } }{ p } =\cfrac { 2p }{ 2 } =2$$
  • Question 6
    1 / -0
    Imagine a new planet having the same density as that of the earth but it is $$3$$ times bigger than the earth is size. If the acceleration due to gravity on the surface of the earth is g and that on the new planet is g', then what is the value of g'/g?
    Solution
    let $${ g }^{ \prime  }$$ and $$g$$ is the acceleration due to gravity at the new planet and at Earth.
    $$\cfrac { { g }^{ \prime  } }{ g } =\cfrac { \cfrac { G{ M }^{ \prime  } }{ { R }^{ \prime 2 } }  }{ \cfrac { GM }{ { R }^{ 2 } }  } =\left( \cfrac { { M }^{ \prime  } }{ M }  \right) \times \left( \cfrac { R }{ { R }^{ \prime  } }  \right) ^{ 2 }=\left( \cfrac { { V }^{ \prime  }P }{ VP }  \right) \left( \cfrac { R }{ { R }^{ \prime  } }  \right) ^{ 2 }\quad \left[ { V }^{ \prime  }=3V\\ =>p.\cfrac { 4 }{ 3 } \pi { R }^{ \prime 3 }=3p.\cfrac { 4 }{ 3 } \pi { R }^{ 3 }\\ =>{ R }^{ \prime 3 }=3{ R }^{ 3 }\\ =>\cfrac { { R }^{ \prime  } }{ R } =(3{ ) }^{ \cfrac { 1 }{ 3 }  } \right] \\ =\cfrac { \cfrac { 4 }{ 3 } \pi { R }^{ \prime 3 } }{ \cfrac { 4 }{ 3 } \pi { R }^{ 3 } } \times \cfrac { { R }^{ 2 } }{ { R }^{ \prime 2 } } \\ =\cfrac { { R }^{ \prime  } }{ R } \\ =(3{ ) }^{ \cfrac { 1 }{ 3 }  }$$

  • Question 7
    1 / -0
    KEPLER'S LAWS
    A satellite is in an elliptic orbit around the earth with aphelion of $${6R}_{E}$$ and perihelion of $${2R}_{D}$$ where $${R}_{E}$$ is the radius of the earth. The eccentricity of the orbit is:
    Solution
    Here, $${r}_{A}={6R}_{E}, {r}_{P}={2R}_{E}$$
    The eccentricity of the orbit is
    $$e=\dfrac{{r}_{A}-{r}_{P}}{{r}_{A}+{r}_{P}};$$

    $$ e=\dfrac{{6R}_{E}-{2R}_{E}}{{6R}_{E}+{2R}_{E}}$$

    $$=\dfrac{4}{8}=\dfrac{1}{2}$$

  • Question 8
    1 / -0
    In a hypothetical uniform and spherical planet of mass $$M$$ and radius $$R$$, a tunnel is dug radially from its surface to its centre as shown. The minimum energy required to carry a unit mass from its centre to the surface is $$kmgR$$. Find value of $$k$$. Acceleration due to gravity at the surface of the planet is $$g$$.

    Solution

  • Question 9
    1 / -0
    If a mass $$m$$ is placed in the vicinity of another mass $$M$$, It experiences a gravitational force of attraction. However, if these two masses are at a very large distance, the force of attraction is negligible. The gravitation potential is the amount of work done per unit mass in bringing it slowly from infinity to some finite distance from $$M$$. Hence, gravitational potential is a state function rather than a path function.
    One application of this potential is in finding the escape velocity of a body from earth. As we know the gravitation potential energy associated with mass $$m$$ on earth surface is $$-\dfrac {GMm}{R}$$. The mechanical energy conservation gives $$\underline {V_{escape} = \sqrt {\dfrac {2GM}{R}}}$$ or $$V_{escape} = \sqrt {2gR}$$. An inquisitive mind decides to get the result by kinematics, considering a particle following curvilinear path from surface to infinity, gravity changes with height as $$g' = g\left (1 + \dfrac {h}{R}\right )^{-2}$$ or $$v\dfrac {dv}{dh} = -g\left (1 + \dfrac {h}{R}\right )^{-2}$$. Solution of the equation within limits $$h = 0$$ to $$h = \infty$$ gives $$\underline {V_{escape} = \sqrt {2gR}}$$.
    For most of the objects, either point masses or objects of finite dimension, the variation of potential in space exhibits symmetric behaviour. In case of spherical objects of uniform density, the locus of equipotential points is a spherical shell of any given radius. Hence, this potential field is symmetric about all three axes passing through the centre of spherical object. Consider a somewhat complicated objects as shown in the adjacent figure.
    The figure shows a solid cube of edge length $$10\ cm$$. The origin is the centre of cube as shown. Eight spherical cavities are formed in this cube, each having a radius of $$1\ cm$$ and centers at $$(\pm 2\ cm, \pm 2\ cm,\pm 2\ cm$$). This figure shows wide range of equipotential surfaces / curves.
    If a particle is thrown with escape velocity $$(v_{e})$$ from surface of earth, which of the following is the correct energy equation?

  • Question 10
    1 / -0

    A rocket is launched normal to the surface of the earth, away from the sun, along the line joining the sun and the earth. The sun is $$3 \times {10^5}$$ times heavier than the earth and is at a distance $$2.5 \times {10^4}$$ times larger than the radius of the earth. The escape velocity from earths gravitational field is $${{\rm{V}}_{\rm{c}}}\;{\rm{ = 11}}{\rm{.2km}}{{\rm{s}}^{{\rm{ - 1}}}}$$ . The minimum initial $$\left( {{{\rm{V}}_{\rm{c}}}} \right)\;$$ required for the rocket to be able to leave the sun-earth system is closest to (Ignore the rotation and revolution of the earth and the presence of any other planet)

    Solution
    $$\dfrac { 1 }{ 2 } { mV }_{ e }^{ 2 }-\dfrac { { GM }_{ e }m }{ { R }_{ e } } -\dfrac { { GM }_{ e }m\times 3\times { 10 }^{ 5 } }{ 2.5\times { 10 }^{ 4 }{ R }_{ e } } =0$$
    $$\dfrac { { Ve }^{ 2 } }{ 2 } =\dfrac { { GM }_{ m } }{ { R }_{ e } } \left[ \dfrac { 1+3\times { 10 }^{ 5 } }{ 2.5\times { 10 }^{ 4 } }  \right] $$
    $${ V }_{ e }=\sqrt { \dfrac { 13\left( { 2GM }_{ e } \right)  }{ { R }_{ e } }  } =\sqrt { 13 } \times 11.2\approx 42$$
    The minimum initial $${ V }_{ s }=42km/s$$.
Self Studies
User
Question Analysis
  • Correct -

  • Wrong -

  • Skipped -

My Perfomance
  • Score

    -

    out of -
  • Rank

    -

    out of -
Re-Attempt Weekly Quiz Competition
Self Studies Get latest Exam Updates
& Study Material Alerts!
No, Thanks
Self Studies
Click on Allow to receive notifications
Allow Notification
Self Studies
Self Studies Self Studies
To enable notifications follow this 2 steps:
  • First Click on Secure Icon Self Studies
  • Second click on the toggle icon
Allow Notification
Get latest Exam Updates & FREE Study Material Alerts!
Self Studies ×
Open Now