Physics Gravitation

The angular momentum of a planet of mass M moving around the sun in an elliptical orbit is L. The magnitude of the areal velocity of the planet is:

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Contributor-Level 10

A = Area swept ⇒ dA/dt = (1/2)r² (dθ/dt) = (1/2) (Mr²ω)/M = L / (2M)

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A particle of mass m moves in a circular orbit under the central potential field, V(r) = -C/r, where C is a constant. The correct radius-velocity graph of the particle's motion is;

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Contributor-Level 10

U = mV (r) = -Cm/r

F = -dU/dr = -Cm/r² ⇒ The force which provides required centripetal force

⇒ mv²/r = Cm/r² ⇒ r = C/v²

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A geostationary satellite is orbiting around an arbitrary planet 'P' at a height of 11R above the surface of 'P', R being the radius of 'P'. The time period of another satellite in hours at a height of 2R from the surface of 'P' is ---------. 'P' has the time period of 24 hours.

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Contributor-Level 10

(T / 24) = (3/12)^ (3/2) ⇒ T = 3 hours

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The time period of satellite in a circular orbit of radius R is T. The period of another satellite in a circular orbit of radius 9R is:

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Raj Pandey

Contributor-Level 9

T² ∝ R³
⇒ (T? /T? )² = (R? /R? )³ ⇒ T? = 9³/² T = 27T

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The radius in kilometre to which the radius of earth (R = 6400 km) to be compressed so that the escape velocity is increased 10 times is .

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The formula for escape velocity (v_e) is v_e = √ (2GM/R).
According to the question, the new escape velocity (v_e') from a new radius R' is related to the original escape velocity by 10v_e' = v_e.
10 * √ (2GM/R') = √ (2GM/R)
Squaring both sides:
100 * (2GM/R') = (2GM/R)
100/R' = 1/R
R' = R/100

If R is the radius of Earth (6400 km), then:
R' = 6400 km / 100 = 64 km

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The value of the acceleration due to gravity is g₁ at a height h = R/2 (R = radius of the earth) from the surface of the earth. It is again equal to g₁ at a depth d below the surface of the earth. The ratio (d/R) equals:

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Contributor-Level 10

dW? = Eqdx
∫ dU? = ∫ kQ/x² dxq
W? = kQq (-1/x)|? ^ (R+y)
W? = kQq (y)/ (R) (R + y)
W? + W? = 1/2 mv²
V² = 2/m (kQqy/ (R) (R+y) + mgy)
V² = 2y (kQq/ (m (R) (R+y) + g) ; k = 1/ (4πε? )

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A body is moving in a low circular orbit about a planet of mass M and radius R. The radius of the orbit can be taken to be R itself. Then the ratio of the speed of this body in the orbit to the escape velocity from the planet is :

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Contributor-Level 10

Escape velocity = √ (2GM/R) = v? f
Orbital speed = √ (GM/R) = V?
V? /V? = 1/√2

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Two planets have masses M and 16M and their radii are a and 2a, respectively. The separation between the centres of the planets is 10a. A body of mass m is fired from the surface of the larger planet towards the smaller planet along the line joining their centres. For the body to be able to reach at the surface of smaller planet, the minimum firings speed needed is:

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On the x-axis and at a distance x from the origin, the gravitational field due to a mass distribution is given by Ax/(x²+a²)³/² in the x-direction. The magnitude of gravitational potential on the x-axis at a distance x, taking its value to be zero at infinity is:

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Contributor-Level 10

g = Ax / (x² + a²)³?²
⇒ ∫? dV = -∫?^∞ gdx
⇒ 0 - V = -∫?^∞ [Ax / (a² + x²)³?²] dx
Let, a² + x² = t²
⇒ 2xdx = 2tdt
⇒ xdx = tdt
⇒ V = ∫?^∞ (Atdt / t³) ⇒ [-A / t]?^∞ ⇒ [-A / √(a² + x²)]?^∞
⇒ V = A / √(a² + x²)

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The mass density of a planet of radius $R$ varies with the distance $r$ from its centre as $ρ (r) = ρ_{0} (1 - \frac{r^{2}}{R^{2}})$ . Then the gravitational field is maximum at:

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