Physics Gravitation

Contributor-Level 10

$\frac{E_{A}}{E_{B}} = \frac{- \frac{G M m_{A}}{2 * 3 r}}{\frac{- G M m_{B}}{2 * 4 r}} = \frac{m_{A}}{m_{B}} * \frac{4}{3} = \frac{4}{3} * \frac{4}{3} = \frac{16}{9}$

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The height of any point P above the surface of earth is equal to diameter of earth. The value of acceleration due to gravity at point P will be: (Given g = acceleration due to gravity at the surface of earth.

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Contributor-Level 10

$g^{1} = \frac{G M}{{(R + h)}^{2}}$

$g^{1} = \frac{g}{{(1 + \frac{h}{R})}^{2}}$

Given h = D = 2R

$g^{1} = \frac{g}{{(1 + \frac{2 R}{R})}^{2}}$

$g^{1} = \frac{g}{9}$

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Two satellites S₁ and S₂ are revolving in circular orbits around a planet with radius R₁ = 3200km and R₂ = 800km respectively. The ratio of speed of satellite S₁ to the speed of satellite S₂ in their respective orbits would be $\frac{1}{x}$ where x =………….

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Contributor-Level 10

M₁ ->Mass of satellite (1)

M₂ -> Mass of satellite (2)

M_P -> Mass of planet

Now NLM (2) on S₁

_{$\frac{G M_{1} M_{P}}{R_{1}^{2}} = \frac{m_{1} v_{1}^{2}}{R_{1}}$}

Similarly v₂ = $\sqrt[]{\frac{G M_{P}}{R_{2}}}$

$\frac{v_{1}}{v_{2}} = \sqrt[]{\frac{R_{2}}{R_{1}}} = \sqrt[]{\frac{800}{3200}} = \frac{1}{2} = \frac{1}{x}$

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An object of mass 1 kg is taken to a height form the surface of earth which is equal to three times that radius of earth. The gain in potential energy of the object will be

[If, g = 10ms^-2 and radius of earth = 6400km]

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Payal Gupta

Contributor-Level 10

m = 1kg ; $Δ U = \frac{- G m_{e} m}{(R + h)} - (\frac{G m_{e} m}{R})$

$Δ U = \frac{G M m}{R} - \frac{G M m}{R + h}$

$= m g_{0} R - m g_{0} \frac{R}{(1 + \frac{h}{R})} = m g_{0} {1 - \frac{1}{(1 + \frac{3 R}{R})}}$

$= m g_{0} R (\frac{3}{4})$

$= \frac{3}{4} * 1 * 10 * 6400 k m$

$= 48000 * 1 0^{3} J .$

$= 48 M J .$

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Given below are two statements:

Statement I: The law of gravitation holds good for any pair of bodies in the universe.

Statement II: The weight of any person becomes zero when the person is at the centre of the earth.

In the light of the above statements, choose the correct answer from the option given below:

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Physics Gravitation Gravitation

Contributor-Level 10

The law of gravitation holds good for any pair of bodies in the universe.

This is correct statement.

The weight of any person becomes zero when the person is at the centre of the Earth.

Weight = mg

= m * 0 [g = 0 at the centre of the Earth]

= 0

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A box weighs $196 N$ on a spring balance at the north pole. Its weight recorded on the same balance if it is shifted to the equator is close to (Take $g = 10 {m s}^{- 2}$ at the north pole and the radius of the earth $= 6400 k m$ )

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Contributor-Level 10

$W_{equator} = W_{pole} - m ω^{2} R$

$\begin{array}{r} = 196 - 19.6 * {(\frac{2 π}{24 * 60 * 60})}^{2} * 6400 * 10^{3} \\ = 195.33 N \end{array}$

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Three identical particles A, B and C of mass 100kg each are placed in a straight line with AB = BC = 13m. The gravitational force on a fourth particle P of the same mass is F, when placed at a distance 13m from the particle B on the perpendicular bisector of the line AC. The value of F will be approximately:

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Contributor-Level 10

F on P = $\frac{}{^{}} \frac{}{^{}}$ $\frac{G M M}{r^{2}} + \sqrt[]{2} \frac{G M M}{{(\sqrt[]{2} r)}^{2}}$

$F o n P = \frac{G * 1 0^{4}}{1 3^{2}} (1 + \frac{1}{\sqrt[]{2}}) \approx 100 G$

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Two satellites A and B, having masses in the ration 4 : 3, are revolving in circular orbits of radii 3r and 4r respectively around the earth. The ratio of total mechanical energy of A to B is:

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Contributor-Level 10

$\frac{E_{A}}{E_{B}} = \frac{- \frac{G M m_{A}}{2 * 3 r}}{\frac{- G M m_{B}}{2 * 4 r}} = \frac{m_{A}}{m_{B}} * \frac{4}{3} = \frac{4}{3} * \frac{4}{3} = \frac{16}{9}$

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An object is taken to a height above the surface of earth at a distance $\frac{5}{4} R$ from the centre of the earth. Where radius of earth, R = 6400km. The percentage decrease in the weight of the object will be

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Payal Gupta

Contributor-Level 10

As we know that $g_{P} = g {(\frac{R}{R + h})}^{2} = g {(\frac{R}{\frac{5}{4} R})}^{2} = \frac{16 g}{25}$

$\Rightarrow Δ g = g - g_{P} = \frac{9 g}{25}$

The percentage decrease in the weight of the object = $\frac{Δ g}{g} * 100 = 36 %$

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2 months ago

If the radius of earth shrinks by 2% while its mass remains same. The acceleration due to gravity on the earth's surface will approximately:

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