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Physics Gravitation

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Physics Gravitation Gravitation

A solid sphere of radius R gravitationally attracts a particle placed at 3R form its centre with a force F₁,. Now a spherical cavity of radius $(\frac{R}{2})$ is made in the sphere (as shown in figure) and the force becomes F₂. The value of F₁: F₂ is:

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Vishal Baghel

Contributor-Level 10

$F_{1} = \frac{G M m}{9 R^{2}}$

$F_{2} = \frac{G M m}{9 R^{2}} - \frac{G \frac{M}{8} m}{{(\frac{5 R}{2})}^{2}} = \frac{G M m}{R^{2}} (\frac{1}{9} - \frac{1}{50})$

$\frac{F_{1}}{F_{2}} = \frac{1}{1 - \frac{9}{50}} = \frac{50}{41}$

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Physics Gravitation Class 11th

The initial velocity vi required to project a body vertically upward from the surface of the earth to reach a height of 10R, where R is the radius of the earth, may be described in terms of escape velocity v_e such that $v_{i} = \sqrt[]{\frac{x}{y}} * v_{e} .$ The value of x will be………

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Payal Gupta

Contributor-Level 10

Using conservation of energy :

$\frac{1}{2} m v_{i}^{2} - \frac{G m m}{R} = 0 - \frac{G m m}{11 R}$

$v_{i}^{2} = \frac{20}{11} \frac{G m}{R}$

Escape velocity is defined as $\sqrt[]{\frac{2 G M}{R}}$

$v_{i} = \sqrt[]{\frac{10}{11} v_{e}}$

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Physics Gravitation Gravitation

The minimum energy required to launch a satellite of mass $m$ from the surface of earth of mass $M$ and radius $R$ in a circular orbit at an altitude of $2 R$ from the surface of the earth is:

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alok kumar singh

Contributor-Level 10

Apply energy conservation,

$U_{i} + K_{i} = U_{f} + K_{f}$

$\Rightarrow - \frac{G M m}{R} + K_{i} = - \frac{G M m}{3 R} + \frac{1}{2} m v^{2}$

$\Rightarrow - \frac{G M m}{R} + K_{i} = - \frac{G M m}{3 R} + \frac{1}{2} * m * \frac{G M}{3 R}$

$\Rightarrow K_{i} = - \frac{1}{6} \frac{G M m}{R} + \frac{G M m}{R}$

$K_{i} = \frac{5}{6} \frac{G M m}{R}$

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Physics Gravitation Gravitation

A parallel plate capacitor is charged by connecting it to a battery through a resistor. If $I$ is the current in the circuit, then in the gap between the plates:

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alok kumar singh

Contributor-Level 10

According to modified Ampere's law

$? B . d I = μ_{0} (I_{C} + I_{D})$

For Loop $L_{1} I_{C} \neq 0$ and $I_{D} = 0$

For Loop $L_{2} I_{C} = 0$ and $I_{D} \neq 0$

Due to $KCL I_{C} = I_{D}$

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Physics Gravitation Gravitation

The satellites revolve around a planet in coplanar circular orbits in anticlockwise direction. Their period of revolutions are 1 hour and 8 hour respectively. The radius of the orbit of nearer satellite is 2 * 10³ km. The angular speed of the farther satellite as observed from the nearer satellite at the instant when both the satellites are closest is $\frac{π}{x} r a d h^{- 1}$ where x is__________.

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alok kumar singh

Contributor-Level 10

For A satellite T₁ = 1hour

$S o ω_{1} = 2 π r e d / h o u r$

for B satellite T₂ = 8 hour

given R₁ = 2 * 10³ Km

Relative $ω = \frac{V_{1} - V_{2}}{R_{2} - R_{1}} = \frac{2 π * 1 0^{3}}{6 * 1 0^{3}}$

$\Rightarrow \frac{π}{3}$ rad/hour

x = 3

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Physics Gravitation Gravitation

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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Physics Gravitation Class 11th

The percentage decrease in the weight of a rocket, when taken to a height of 32 km above the surface of earth will, be: (Radius of earth will, be

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Payal Gupta

Contributor-Level 10

$g^{1} = g (1 - \frac{2 h}{R}) = g (1 - 2 * \frac{32}{6400})$

$g^{1} = \frac{99 g}{100} = 0.99 g$

% decrease is wt = $\frac{g - g^{1}}{g} * 100 = 1 %$

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Physics Gravitation Gravitation

The mass of a planet is $\frac{1}{10}$ th that of the earth and its diameter is half that of the earth. The acceleration due to gravity on that planet is:

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alok kumar singh

Contributor-Level 10

$g^{'} = \frac{G M^{'}}{R^{' 2}} = \frac{G M}{10 {(\frac{R}{2})}^{2}}$

$= \frac{4}{10} \frac{G M}{R^{2}} = 0.4 * 9.8$

$= 3.92 m s^{- 2}$

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Physics Gravitation Gravitation

A body is projected vertically upwards from the surface of earth with a velocity sufficient enough to carry it to infinity. The time taken by it to earth reach height h is________

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Vishal Baghel

Contributor-Level 10

Using conservation of energy :

$? \frac{? G M m}{R} + \frac{1}{2} m {(\sqrt[]{\frac{2 G M}{R}})}^{2} = \frac{? G M m}{r} + \frac{1}{2} m v^{2}$

$? \frac{1}{2} m v^{2} = \frac{G M m}{r}$

$? ?_{R}^{R + h} r^{1 / 2} d r = \sqrt[]{2 G M} ?_{0}^{t} d t$

$? \frac{2}{3} R^{3 / 2} [{(1 + \frac{h}{R})}^{3 / 2} ? 1] = \sqrt[]{2 G M} t$

$t = \frac{1}{3} \sqrt[]{\frac{2 R}{g}} [{(1 + \frac{h}{R})}^{3 / 2} ? 1]$

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Physics Gravitation Gravitation

A person whose mass is 100kg travels from Earth to Mars in a spaceship. Neglect all other object in sky and take acceleration due to gravity on the surface of the Earth and mars as 10 m/s² and 4m/s² respectively. Identify from the below figures, the curve that fits best for the weight of the passenger as a function of time.

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alok kumar singh

Contributor-Level 10

W₁ = 1000 N ->Weight of person at surface of earth W₂ = 400 N Þ Weight of person at surface of Mars. There will be a neutral point -N where weight will be zero because at this point net gravitational field due to earth and mars is zero.

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