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Physics Gravitation

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3 months ago

Physics Gravitation Class 11th

The planet Mars has two moons, if one of them has a period 7 hours, 30 minutes and an orbital radius of 9.0 * 10³ km. Find the mass of Mars.
{Given 4π²/G = 6 * 10¹¹ N⁻²kg²}

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Vishal Baghel

Contributor-Level 10

T² = (4π²r³)/GM
⇒ M = (4π²r³)/ (G T²)
⇒ M = 6 * 10¹¹ * (9 * 10? )³ / (450 * 60)²
= 6.48 x 10²³ kg
(Note: There is a calculation error in the source image. Using the formula and given values: M = (6e11 * (9e6)^3) / (27000)^2 = 6e23 kg)

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Physics Gravitation Class 11th

Suppose two planets (spherical in shape) of radii R and 2R, but mass M and 9M respectively have a centre to centre separation 8R as shown in the figure. A satellite of mass 'm' is projected from the surface of the planet of mass 'M' directly towards the centre of the second planet. The minimum speed 'v' required for the satellite to reach the surface of the second planet is √(aGM/(7R)) then the value of 'a' is _______. [Given: The two planets are fixed in their position]

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Vishal Baghel

Contributor-Level 10

We have to find the point where the gravitational field must be zero.
EG = 0
GM/x² = G (9M)/ (8R-x)²
1/x² = 9/ (8R-x)²
8R - x = 3x => x = 2R
Potential at A (surface of first planet), VA = -GM/R - G (9M)/7R = -16GM/7R
Potential at point x, Vx = -GM/x - G (9M)/ (8R-x) = -GM/2R - G (9M)/6R = -2GM/R
ΔV = Vx - VA = -2GM/R - (-16GM/7R) = (-14+16)GM/7R = 2GM/7R
Using conservation of mechanical energy
ΔKE = ΔU = mΔV
½mv² = m (2GM/7R)
v² = 4GM/7R
v = √ (4GM/7R) => a = 4

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3 months ago

Physics Gravitation Class 11th

Consider a planet in some solar system which has a mass double the mass of earth and density equal to the average density of earth. If the weight of an object on earth is W, the weight of the same object on that planet will be :

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alok kumar singh

Contributor-Level 10

g = (4/3)πRρG . (i) (ρ = density)
Now ρ = M/V = M/ (4/3)πR³)
R³ = M/ (4/3)πρ) => R ∝ (M/ρ)¹/³
From equation (i)
g ∝ Rρ ∝ (M/ρ)¹/³ρ = M¹/³ρ²/³
For planet, M' = 2M, ρ' = ρ
g'/g = (M'/M)¹/³ (ρ'/ρ)²/³ = (2)¹/³ (1)²/³ = 2¹/³
W' = mg' = m (2¹/³g) = 2¹/³ (mg) = 2¹/³W

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Physics Gravitation Class 11th

The minimum and maximum distances of a planet revolving around the Sun are x? and x?. If the minimum speed of the planet on its trajectory is v?, then its maximum speed will be :

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Physics Gravitation 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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Physics Gravitation 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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Physics Gravitation 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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3 months ago

Physics Gravitation Class 11th

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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Vishal Baghel

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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3 months ago

Physics Gravitation Class 11th

Assume that a tunnel is dug along a chord of the earth, at a perpendicular distance (R/2) from the earth's centre, where 'R' is the radius of the Earth. The wall of the tunnel is frictionless. If a particle is released in this tunnel, it will execute a simple harmonic motion with a time period

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Payal Gupta

Contributor-Level 10

F = m E cos θ

$= m (\frac{G M}{R^{3}} * r) * \frac{x}{r}$ $m a = \frac{m g}{R} * \Rightarrow a = \frac{g}{R} x$

$\Rightarrow T = 2 π \sqrt[]{\frac{R}{g}}$

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3 months ago

Physics Gravitation Class 11th

Find the gravitational force of attraction between the ring and sphere as shown in the diagram where the plane of the ring is perpendicular to the joining the centres. If $\sqrt[]{8} R$ is the distance between the centres of a ring (of mass 'm') and a sphere (mass 'M') where both have equal radius 'R'.

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Payal Gupta

Contributor-Level 10

Method – I:

$F = M_{s p h e r e} E_{r i n g} = M * \frac{G m x}{{(R^{2} + x^{2})}^{\frac{3}{2}}} = M * \frac{G m x \sqrt[]{8} R}{{(R^{2} + 8 R^{2})}^{\frac{3}{2}}} = \frac{\sqrt[]{8} G M m}{27 R^{2}}$

Method – II:

$F = \int d F c o s θ = c o s θ \int (d m) \frac{G M}{(R^{2} + 8 R^{2})} = \frac{\sqrt[]{8} G M m}{27 R^{2}}$

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