Physics Gravitation

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New answer posted•a year ago

Physics Gravitation Gravitation

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:

2 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

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

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New answer posted•a year ago

Physics Gravitation Class 11th

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.

2 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

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

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

Given h = D = 2R

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

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

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New answer posted•a year ago

Physics Gravitation Class 11th

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 =………….

3 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

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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New answer posted•a year ago

Physics Gravitation Class 11th

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]

13 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

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

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

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

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

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

$= 48000 \times 1 0^{3} J .$

$= 48 M J .$

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New answer posted•a year ago

Physics Gravitation Class 11th

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:

4 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

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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New answer posted•a year ago

Physics Gravitation Gravitation

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 $= 6400 k m$ )

2 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

$W_{equator} = W_{pole} - m ω^{2} R$

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

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New answer posted•a year ago

Physics Gravitation Class 11th

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:

11 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

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

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

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New answer posted•a year ago

Physics Gravitation Class 11th

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:

6 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

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