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2 months ago

Overview Class 11th

A satellite is moving around a planet of radius R in a circular orbit of radius r. The speed of the satellite is v. Find the acceleration due to gravity on the surface of the planet.

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Vishal Baghel

Contributor-Level 10

v = √ (GM/r) (orbital speed)
⇒ GM = v²r ⇒ (GM/R²) = (v²r/R²) ⇒ g = (v²r/R²)

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New answer posted

2 months ago

Overview Class 11th

In free space, two particles of mass m each are initially both at rest at a distance a from each other. They start moving towards each other due to their mutual gravitational attraction. The time after which the distance between them has reduced to a/2 is

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Vishal Baghel

Contributor-Level 10

Let the origin be at the CM of the particles, let the initial positions of the particles be x = a/2 and x = -a/2 and let the instantaneous positions of the particles be x = r and x = -r
Let the instantaneous velocity of each particle be v
Let the time after which the distance between the particles has reduced to a/2 be T
Then, for the particle that was initially at x = -a/2,
(Gm²/ (2r)²) = -m (vdv/dr) ⇒ (Gm/r²) = - (4vdv/dr) ⇒ -4∫vdv = Gm∫ (dr/r²)
[v is negative because the velocity is towards the –X direction]
dr/dt = -√ (Gm/2) (1/r - 2/a)¹? ²
⇒ ∫ (a/2)^ (a/4) (r/√ (a-2r)dr = -√ (Gm/2a) ∫? dt
⇒ ∫ (a/2)^ (a/4

...more

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New answer posted

2 months ago

Overview Class 11th

Two stars of masses m and 2m at a distance d rotate about their common centre of mass in free space. The

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alok kumar singh

Contributor-Level 10

$m ω^{2} . \frac{2 d}{3} = \frac{G . m . 2 m}{d^{2}} \Rightarrow ω = \sqrt[]{\frac{3 G m}{d^{3}}}$

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

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New answer posted

2 months ago

Overview Class 11th

Four identical particles of equal masses 1kg made to move along the circumference of a circle of radius 1m under the actin of their own mutual gravitational attraction. The speed of each particle will be:

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alok kumar singh

Contributor-Level 10

$F_{1} c o s 45 ° + F_{2} c o s 45 ° + F_{3} = \frac{m v^{2}}{r}$

$\Rightarrow \frac{G m^{2}}{{(\sqrt[]{2} r)}^{2}} . \frac{1}{\sqrt[]{2}} + \frac{G m^{2}}{{(\sqrt[]{2} r)}^{2}} . \frac{1}{\sqrt[]{2}} + \frac{G m^{2}}{{(2 r)}^{2}} = \frac{m v^{2}}{r}$

$\Rightarrow v = \sqrt[]{\frac{G m}{4 r} (2 \sqrt[]{2} + 1)}$

Putting m = 1 kg and r = 1 m,

$v = \frac{1}{2} \sqrt[]{G (1 + 2 \sqrt[]{2})}$

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New answer posted

2 months ago

Overview Class 11th

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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alok kumar singh

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

2 months ago

Overview 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.

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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

2 months ago

Overview Class 11th

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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alok kumar singh

Contributor-Level 10

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

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New answer posted

2 months ago

Overview Class 11th

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

2 months ago

Overview Class 11th

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

2 months ago

Overview Class 11th

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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alok kumar singh

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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