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In the reported figure of earth, the value of acceleration due to gravity is same at point A and C but it is smaller than that of its value at point B (surface of the earth). The value of OA: AB will be x : y. The value of x is…………

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Contributor-Level 10

$\frac{G M}{{(3 R / 2)}^{2}} = \frac{G M}{R^{3}} * r$

$\Rightarrow O A = \frac{4 R}{9} = r$

$A B = R - \frac{4 R}{9} = \frac{5 R}{9} \Rightarrow O A : A B = 4 : 5 = x : y \Rightarrow x = 4$

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2 months ago

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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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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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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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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2 months ago

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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Contributor-Level 10

According to question, we can write

$d = \sqrt[]{2 h R} \Rightarrow \frac{d_{2}}{d_{1}} = \sqrt[]{\frac{h_{2}}{h_{1}}} \Rightarrow h_{2} = {(\frac{d_{2}}{d_{1}})}^{2} h_{1} = {(\frac{2}{1})}^{2} * 125 = 500 m$

Increment in height of tower = h₂ – h₁ = 500 – 125 = 375 m

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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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Contributor-Level 10

Kindly go through the solution

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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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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

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

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

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Contributor-Level 10

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

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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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