NEET Physics Questions: Gravitation

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The gravitational force Fg between two objects does not depend on




The gravitational force \( F_g \) between two objects is given by the formula \( F_g = G \frac{m_1 m_2}{r^2} \), where \( G \) is the gravitational constant, \( m_1 \) and \( m_2 \) are the masses of the objects, and \( r \) is the distance between them. The force depends on the product of the masses, not their sum. Therefore, the correct option is 'sum of the masses'.
Two sphere of mass $m_1$ and $m_2$ are situated in air and the gravitational force between them is F. The space around the masses is now filled with liquid of specific gravity 3. The gravitational force will now be




Gravitational force does not depend upon the medium
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 $
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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




No explanation available.
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




No explanation available.
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 } $
The radius of the earth is 6400 km and $g=10ms^{-2}$. In order that a body of 5 kg weights zero at the equator, the angular speed of the earth is = rad/sec




for condition of weight lessness at equator $ \omega = \sqrt { g / R } $ $ = \sqrt { 10 \over 6400 \times 10^3 } = { 1 \over 800} rad /sec $
The time period of a simple pendulum on a freely moving artificial satellite is




Time peripd of simple pendylym $ T = 2 \pi \sqrt { l /g } $ In artificial satellite g ' = 0 $ \therefore T = \infty $
The value of g on he earth surface is $980 cm/sec^2$. Its value at a height of 64 km from the earth surface is …………..$cm5^{–2} $




$ { g' \over g } = \left( { R \over (R+h) } \right)^2 = \left( {6400 \over 6400+64} \right) \Rightarrow g' = 960.400 ms^{-2} $