Home
Class 10
PHYSICS
Gravitational force, F=Gm(1)m(2)//r....

Gravitational force, `F=Gm_(1)m_(2)//r.`

Answer

Step by step text solution for Gravitational force, F=Gm_(1)m_(2)//r. by PHYSICS experts to help you in doubts & scoring excellent marks in Class 10 exams.

Doubtnut Promotions Banner Mobile Dark
|

Topper's Solved these Questions

  • TRUE OR FALSE

    NAVNEET PUBLICATION - MAHARASHTRA BOARD|Exercise Space Missions|4 Videos
  • TOWARDS GREEN ENERGY

    NAVNEET PUBLICATION - MAHARASHTRA BOARD|Exercise MCQS BASED ON PROJECTS|7 Videos
  • TRUE/FALSE

    NAVNEET PUBLICATION - MAHARASHTRA BOARD|Exercise ASSIGNMENT|2 Videos

Similar Questions

Explore conceptually related problems

A mass M is split into two parts m and (M-m) which are then separated by certain distance. Find ratio (m/M) to maximise the gravitational force F=(Gm(M-m))/(r^(2)) between the parts. Here G = gravitational constant and r is the distance between m and (M-m).

Explain with a diagram : Kepler's three laws. Hence show that gravitational force, F prop (1)/(r^(2)) (in the usual notation).

Knowledge Check

  • Two stars of mass m_(1) and m_(2) are parts of a binary star system. The radii of their orbits are r_(1) and r_(2) respectivey, measured from the centre of mass of the system. The magnitude of gravitational force m_(1) exerts on m_(2) is

    A
    `(m_(1)m_(2)G)/((r_(1)+r_(2))^(2))`
    B
    `(m_(1)G)/((r_(1)+r_(2))^(2))`
    C
    `(m_(2)G)/((r_(1)+r_(2))^(2))`
    D
    `(G(m_(1)+m_(2)))/((r_(1)+r_(2))^(2))`
  • Supposing Newton's law of gravitation for gravitation force F_(1) and F_(2) between two masses m_(1) and m_(2) at positions r_(1) and r_(2) read F_(2)=-F_(2)=(r_(12))/(r_(12)^(3))GM_(0)^(2)((m_(1)m_(2))/(M_(0)^(2)))^(n) where M_(0) is a constant dimension of mass, r_(12)=r_(1)-r_(2) and n is number. In such a case.

    A
    the acceleration due to gravity on earth will be different for different objects
    B
    none of the three laws kepler will be valid
    C
    only the third law will become invalid
    D
    for `n` negative, an object lighter than water will sink in water.
  • Supposing Newton's law of gravitation for gravitation forces F_(1) and F_(2) between two masses m_(1) and m_(2) at positions r_(1) and r_(2) read where M_(0) is a constant of dimension of mass, r_(12) = r_(1) - r_(2) and n is a number. In such a case,

    A
    the acceleration due to gravity on the earth will be different for different objects
    B
    none of the three laws of Kepler will be valid
    C
    only the third law will become invalid
    D
    for `n` negative, an object lighter than water will sink in water
  • Similar Questions

    Explore conceptually related problems

    Hence show that gravitational force F prop (1)/(r^(2)) ( in the usual notation)

    Derive the dimension of gravitation constant G. Given F_("gravitation")=G(m_(1)m_(2))/(r^(2)) .

    Two concentric shells of masses M_(1) and M_(2) are concentric as shown. Calculate the gravitational force on m due to M_(1) at points P,Q and R .

    A solid sphere of uniform density and radius R applies a gravitational force of attraction equal to F_(1) on a particle placed at P, distance 2R from the centre O of the sphere. A spherical cavity of radius R//2 is now made in the sphere as shown in given figure. The sphere with cavity now applies a gravitational force F_(2) on same particle placed at P. The ratio F_(2)//F_(1) will be

    A solid sphere of uniform density and radius R applies a gravitational force of attraction equal to F_(1) on a particle placed at P, distance 2 R from the centre O of the sphere. A spherical cavity of the radius R//2 is now made in the sphere as shown in the figure. The sphere with the cavity now applies a gravitational force F_(2) on the same particle placed at P. The ratio F_(2)//F_(1) will be