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2 weeks ago

Overview Class 11th

The variation of acceleration due to gravity (g) with distance (r) from the centre of the earth is correctly represented by: (Given R = radius of earth)

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Raj Pandey

Contributor-Level 9

$g \propto r 0 \leq r \leq R$

$g \propto \frac{1}{r^{2}} r > R$

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New answer posted

2 weeks ago

Overview Class 11th

The time period of a satellite revolving around earth in a given orbit is 7 hours. If the radius of orbit is increased to three its previous value, then approximate new time period of the satellite will be

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alok kumar singh

Contributor-Level 10

Kindly go through the solution

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New answer posted

3 weeks 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 are $e$ . Time period of revolution of planet is given by

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alok kumar singh

Contributor-Level 10

$b = a \sqrt[]{1 - e^{2}}$

$\frac{d A}{d t} = \frac{L}{2 m}$

$\frac{A}{T} = \frac{L}{2 m}$

$T = \frac{2 m A}{L} = \frac{2 m π a b}{L} = \frac{2 m π a^{2} \sqrt[]{1 - e^{2}}}{L}$

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New answer posted

3 weeks ago

Overview Class 11th

The diagram showing the variation of gravitational potential of earth with distance from the centre of earth is

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Raj Pandey

Contributor-Level 9

$V_{in} = \frac{- G M}{2 R} [3 - {(\frac{r}{R})}^{2}],$

$V_{s u r f a c e} = \frac{- G M}{R}, V_{out} = \frac{- G M}{r}$

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New answer posted

3 weeks ago

Overview Class 11th

Two satellites A and B of masses 200kg and 400kg are revolving round the earth at height of 600km and 1600km respectively. If T_A and T_B are the time periods of A and B respectively then the value of T_B – T_A :

[Given : radius of earth = 6400km, mass of earth = 6 * 10²⁴kg]

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Vishal Baghel

Contributor-Level 10

$T = 2 π \sqrt[]{\frac{r^{3}}{G M_{e}}}$

$T_{B} - T_{A} = \frac{2 π}{\sqrt[]{G M_{e}}} (r_{B}^{3 / 2} - r_{A}^{3 / 2}) = \frac{2 * 3.14}{\sqrt[]{6.67 * 1 0^{- 11} * 6 * 1 0^{24}}} [{(8 * 1 0^{6})}^{\frac{3}{2}} - {(7 * 1 0^{6})}^{\frac{3}{2}}]$

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New answer posted

3 weeks ago

Overview Class 11th

The minimum and maximum distance of a satellite from the centre of the earth are 2R and 4R, respectively, where R is the radius of earth and M is the mass of earth. The radius of curvature at the point of maximum distance is

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alok kumar singh

Contributor-Level 10

$\frac{1}{2} m V_{a}^{2} - \frac{G M m}{2 x}$

= $\frac{1}{2} m \frac{V_{a}^{2}}{4} - \frac{G M m}{4 x}, V_{a} = \sqrt[]{\frac{2}{3} \frac{G M}{x}}$

ROC at point – B

$R O C = \frac{V_{b}^{2}}{a ⊥} = \frac{\frac{2 G M}{3 R} * \frac{1}{4}}{\frac{G M}{{(4 R)}^{2}}} = \frac{8 R}{3}$

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New answer posted

3 weeks 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

3 weeks 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

3 weeks 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

3 weeks 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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