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The variation of acceleration due to gravity g with distance d from the centre of the earth is best represented by (R is Earth's radius) 

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Important Questions on Gravitation

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Starting from the centre of the earth having radius R,  the variation of g (acceleration due to gravity) is shown by  

 

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The value of gravitational acceleration at a height equal to radius of earth, is 
 
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Time period of second pendulum on a planet, whose mass and diameter are twice that of earth, is 
 
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Calculate angular velocity of earth so that acceleration due to gravity at 60° latitude becomes zero. (Radius of earth is 6400 km gravitational acceleration at poles =10 m s-2 cos60°=0.5

 
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If the mass of the Sun were ten times smaller and the universal gravitational constant were ten times larger in magnitude, which of the following is not correct? 
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Kepler's third law states that the square of the period of revolution (T) of a planet around the sun, is proportional to the third power of the average distance r between sun and planet i.e., T2=Kr3 here K is constant.

If the masses of sun and planet are M and m respectively then as per Newton's law of gravitation force of attraction between them is
F=GMmr2,here G is gravitational constant. The relation between G and K is described as

 

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A planet of mass 'm' moves in an elliptical orbit around an unknown star of mass 'M' such that its maximum and minimum distances from the star are equal to r1 and r2 respectively. The angular momentum of the planet relative to the centre of the star is
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A remote - sensing satellite of earth revolves in a circular orbit at a height of 0.25×106 m above the surface of earth. If earth's radius is 6.38×106 m and g=9.8 m s-2, then the orbital speed of the satellite is