Physics Gravitation

A particle of mass ' m ' is projected with a velocity v = kVe(k < 1) from the surface of the earth.
(Ve = escape velocity)
The maximum height above the surface reached by the particle is:

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Contributor-Level 10

h = R / [ (2gR/V? ²) - 1] = R / [ (V²/K²V? ²) ] = KK² / (1 - K²)

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3 weeks ago

A planet is revolving around the sun in an elliptical orbit. The mass of planet is m, angular momentum of planet about sun is L, and length of semi major axis is a and eccentricity is e. Time period of revolution of planet is given by

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Vishal Baghel

Contributor-Level 10

b = a√1 - e²
dA/dt = L/2m
T = A / (L/2m)
T = 2mA/L = 2mπab/L = 2mπa²√1 - e²/L

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New answer posted

3 weeks ago

A satellite is orbiting just above the surface of the earth with period T. If d is the density of the earth and G is the universal constant of gravitation, the quantity $\frac{3 π}{G d}$ represents:

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Contributor-Level 9

Time period of satellite

$T = 2 π \sqrt[]{\frac{R^{3}}{G M}}$

$= 2 π \sqrt[]{\frac{R^{3}}{G d \frac{4}{3} π R^{3}}}$

$\Rightarrow T = \sqrt[]{\frac{3 π}{G d}}$ $\Rightarrow \frac{3 π}{G d} = T^{2}$

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New answer posted

4 weeks ago

Two bodies of mass m and 9m are placed at a distance R. The gravitational potential on the line joining the bodies where the gravitational field equals zero, will be ( G = gravitational constant):

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Contributor-Level 9

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4 weeks ago

The volume occupied by the molecules contained in 4.5Kg water at STP, if the intermolecular forces vanish away is :

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Contributor-Level 9

$\begin{array}{r} V = (no. of moles) (22.4 litre) \\ = \frac{mass}{molar mass} (22.4 * 10^{- 3} m^{3}) \\ = \frac{4.5 * 10^{3}}{18} * 22.4 * 10^{- 3} m^{3} \\ = 5.6 m^{3} \end{array}$

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4 weeks ago

The escape velocity from the Earth's surface is $v$ . The escape velocity from the surface of another planet having a radius, four times that of Earth and same mass density is:

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Contributor-Level 10

$v_{e} = \sqrt[]{\frac{2 G M}{R}} = \sqrt[]{\frac{2 G}{R} * \frac{4}{3} π R^{3} ρ}$

$= \sqrt[]{\frac{8 π G ρ}{3} R^{2}}$

$\Rightarrow v_{e} \propto R$

$\Rightarrow \frac{v_{e}}{v} = \frac{4 R}{R} \Rightarrow v_{e} = 4 v$

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New answer posted

a month ago

A body of mass 60g experiences a gravitational force of 3.0 N, when placed at a particular point. The magnitude of the gravitational field intensity at that point is :

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Contributor-Level 9

$I_{g} = \frac{F}{m} = \frac{3}{60 * 10^{- 3}} = 50 N / k g$

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New answer posted

a month ago

A satellite is orbiting just above the surface of the earth with period T. If $d$ is the density of the earth and $G$ is the universal constant of gravitation, the quantity $\frac{3 π}{G d}$ represents:

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Physics Gravitation Gravitation

Contributor-Level 10

${\overset{?}{B}}_{P} = {\overset{?}{B}}_{upper wire} \otimes + {\overset{?}{B}}_{semi-circle} ? + {\overset{?}{B}}_{lower-wire} \otimes$

$B_{P} = - \frac{μ_{0} i}{4 π R} + \frac{μ_{0} i}{4 R} - \frac{μ_{0} i}{4 π R}$ $= \frac{μ_{0} i}{4 R} (1 - \frac{2}{π})$ pointing away from the page

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New answer posted

a month ago

Two bodies of mass $m$ and $9 m$ are placed at a distance $R$ . The gravitational potential on the line joining the bodies where the gravitational field equals zero, will be ( $G =$ gravitational constant):

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Contributor-Level 10

Stress $= \frac{F}{A} = \frac{T}{A} = \frac{W}{A}$

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New answer posted

a month ago