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Physics Gravitation

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2 months ago

Physics Gravitation Class 11th

Why is there a negative sign in the total energy formula of a satellite?

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Aadit Singh Uppal

Contributor-Level 10

The minus sign in the formula indicates that:

A gravitational force exists.
The satellite cannot leave the orbit by itself.
A minimum threshold of energy will be required to move the satellite from its existing orbit.

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Physics Gravitation Class 11th

Two planets A and B of equal mass are having their period of revolutions T_A and T_B such that T_A = 2T_B. These planets are revolving in the circular orbits or radii r_A and r_B respectively. Which out of the following would be the correct relationship of their orbits?

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Raj Pandey

Contributor-Level 9

According to kepler's third law of time period –

${(\frac{T_{A}}{T_{B}})}^{2} = {(\frac{r_{A}}{r_{B}})}^{3} \Rightarrow \frac{r_{A}}{r_{B}} = 4^{1 / 3}$

$\Rightarrow r_{4}^{3} = 4 r_{B}^{3}$

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Physics Gravitation 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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2 months ago

Physics Gravitation 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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2 months ago

Physics Gravitation 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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2 months ago

Physics Gravitation 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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2 months ago

Physics Gravitation 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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2 months ago

Physics Gravitation 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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Physics Gravitation Class 11th

The mass density of a planet of radius $R$ varies with the distance $r$ from its centre as $ρ (r) = ρ_{0} (1 - \frac{r^{3}}{R^{3}})$ . Then the gravitational field is maximum at:

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Raj Pandey

Contributor-Level 9

$g = \frac{G M_{in}}{r^{2}}$ , where $M_{in} = \int_{0}^{r} ? ρ 4 π x^{2} \cdot d x$

$\begin{array}{r} M_{in} = 4 π ρ_{0} \int_{0}^{r} ? ? (1 - \frac{x^{3}}{R^{3}}) x^{2} d x = 4 π ρ_{0} (\frac{r^{3}}{3} - \frac{r^{6}}{6 R^{3}}) \\ ∴ g = \frac{4 π G ρ_{0}}{6} (2 r - \frac{r^{4}}{R^{3}}) \end{array}$ $_{}_{}^{}^{}$

For $g$ to be maximum or minimum or constant $\frac{d g}{d r} = 0 \Rightarrow 2 - \frac{4 r^{3}}{R^{3}} = 0 \Rightarrow r = \frac{R}{2^{1 / 3}}$ $\frac{}{} \frac{^{}}{^{}} \frac{}{^{}}$

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2 months ago

Physics Gravitation 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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