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Consider two satellites S₁ and S₂ with periods of revolution 1hr. and 8 hr. respectively revolving around a planet in circular orbits. The ratio of angular velocity of satellite S₁ to the angular velocity of satellites S₂ is:

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Contributor-Level 10

$ω = \frac{2 π}{T} \Rightarrow \frac{ω_{1}}{ω_{2}} = \frac{T_{2}}{T_{1}} = 8$

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New answer posted

3 weeks ago

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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Raj Pandey

Contributor-Level 9

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

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

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

3 weeks ago

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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New answer posted

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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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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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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Raj Pandey

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

a month 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 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 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

If the angular velocity of earth's spin is increased such that the bodies at the equator start floating, the duration of the day would be approximately :
[Take g = 10 ms⁻², the radius of earth, R = 6400*10³ m, Take π = 3.14]

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Contributor-Level 10

g? = g - ω²R ⇒ 0 = g - ω²R ⇒ ω = √ (g/R)

⇒ T = 2π/ω = 2π√ (R/g) = 2 * 3.14 * √ (6400 * 10³)/10)

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New answer posted

a month ago

The angular momentum of a planet of mass M moving around the sun in an elliptical orbit is L. The magnitude of the areal velocity of the planet is:

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Contributor-Level 10

A = Area swept ⇒ dA/dt = (1/2)r² (dθ/dt) = (1/2) (Mr²ω)/M = L / (2M)

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New answer posted

a month ago

A particle of mass m moves in a circular orbit under the central potential field, V(r) = -C/r, where C is a constant. The correct radius-velocity graph of the particle's motion is;

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