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A satellite of the earth is revolveing in a circular orbit with a uniform speed v. If the gravitational force suddennly disappears, the satellite will
Due to inertia to direction
Two particle of equal mass go round a circle of radius r. Under the action of their mutual gravitational force. The speed of each particle is =..................
cenripetal force provided by the gravitational force of attraction between two particites $ \therefore { m \upsilon^2 \over r } = { G (m) (m) \over (2r)^2 } $ $ \therefore \upsilon = {1 \over 2 } \sqrt { Gm \over r } $
The distance of the moon and earth is D the mass of earth is 81 times the mass of moon. At what distance from the center of the earth, the gravitational force will be zero
For will be zero at the point of zero intensity $ x = { \sqrt {m_1} \over \sqrt {m_1} + \sqrt {m_2} } = { \sqrt m \over \sqrt {81m} + \sqrt m }$ $ D = { 9 \over 10 } D $
One can easily “ Weight the earth †by calculating the mass of earth using the formula (in usual notation)
$ mg = { G M_e m \over R e^2 } $ where Me and Re is the mass and radius of the earth respectively $ \therefore M_e = { g \over G } Re^2 $
Three equal masses of m kg each are plced the vertices of an equilateral triangle PQR and a mass of 2m kg is placed at the centroid 0 of the triangle which is at a distance of $\sqrt 2 m$ from each of vertices of triangle. The force in newton. acting on the mass 2m is = ............
Here $ F_{OA} = F_{OB} =F_{OC} = { G (m) (2m ) \over r^2 } $ $ \vec F = \vec F_{OA} + \vec F_{OB} + \vec F _ { OC } $
Which of the following statement about the gravitational constant is true
Two point masses A and B having masses in the ratio 4 : 3 are seprated by a distance of lm. When another point mass c of mass M is placed in between A and B the forces A and C is 1/3rd of the force between band C, Then the distance C form A is = m
Here$ { m_a \over m_b } = { 4 \over 3 } $ $ FAC = { G (m) (mA) \over x^2 } ....(i) $ $ FAC = { G (m) (mB) \over (1-x)^2 } ....(ii) $ According to given problem $ FAC = { 1 \over 4 } FBC $ with the help of eqn (i) and (ii) $ = { G (m) ( mA) \over x^2 } = { 1 \over 3 } { G (m) ( mB ) \over (1-x) ^2 } $ $ \therefore { m_A \over m_B} = { x^2 \over 3(1-x)^2 } \Rightarrow { 4\over 3} = { x^2 \over (1-x)^2 } \Rightarrow 4 = { x^2 \over (1-x)^2} $ $ \Rightarrow 2 = { x \over 1-x } = 2 - 2x = x $ $ = 3x = 2 \therefore x = { 2 \over 3 } m $
The gravitational force between two point masses $m_1$ ans $m_2$ at separation r is given by$ F = G { m_1 m_2 \over r^2 } $. The constant k
As we go from the equator to the poles, the value of g
The value of gravitational acceleration (g) increases as we move from the equator to the poles. This occurs because the Earth is not a perfect sphere but an oblate spheroid, meaning it is slightly flattened at the poles and bulging at the equator. The radius of the Earth is smaller at the poles than at the equator, resulting in a stronger gravitational pull at the poles.
If R is the radius of the earth and g the acceleration due to gravity on the earth’s surface, the mean density of the earth is =
$ g = { GM \over R^2 } and M = { 4 \over 3 } \pi R^3 \rho $ $ \therefore g = { G \over R^2 } . {4 \over 3 } \pi R^3 \rho \Rightarrow \rho = { 3g \over 4 \pi RG } $
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