Which of the following quantities do not depend upon the \(\frac{T}{R}\) orbital radius of the satellite ?
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Solution
According to 3rd law of Kepler
T2αR2
⇒T2=KR3
where K is a constant
Thus\(\frac{T^{2}}{R^{3}}\)does not depends on radius.
The orbital velocity of an artificial satellite in a circular orbit very close to Earth is v. The velocity of a geosynchronous satellite orbiting in a circular orbit at an altitude of 6R from Earth’s surface will be
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Solution
\(v_{1}\alpha \frac{1}{\sqrt{R}},V_{2}\alpha \frac{1}{\sqrt{7R}}\)
\(\frac{v_{2}}{v_{1}}=\frac{1}{\sqrt{7}}\Rightarrow v_{2}=\frac{v_{1}}{\sqrt{7}}=\frac{v}{\sqrt{7}}\)
Escape velocity when a body of mass m is thrown vertically from the surface of the earth is v, what will be the escape velocity of another body of mass 4 m is thrown vertically
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Solution
Escape velocity does not depend upon the mass of the body
The distance of neptune and saturn from the sun is nearly 1013 and 1012 meter respectively.As suming that they move in circular orbits, their periodic times will be in the ratio
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Solution
\(T^{2}\alpha R^{3}(According\: to \: Kepler’s \: law)\)
\(T^{2}\alpha R^{3}(According\: to \: Kepler’s \: law)\)
\(∴\frac{T_{1}^{2}}{T_{2}^{2}}=(10)^{3}\: or\: \frac{T_{1}}{T_{2}}=10\sqrt{10}\)
Directions: Each question contains STATEMENT-1 and STATEMENT-2. Choose the correct answer(ONLY ONE option is correct ) from the following
Statement -1:Gravitational potential is maximum at infinity.
Statement -2 :Gravitational potential is the amount of work done to shift a unit mass from infinity to a given point in gravitational attraction force field.
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Solution
(b)Statement -1 is true, Statement-2 is true;Statement -2 is a correct explanation for Statement-1
Directions: Each question contains STATEMENT-1 and STATEMENT-2. Choose the correct answer(ONLY ONE option is correct ) from the following
Statement -1 : For the planets orbiting around the sun,angular speed, linear speed and K.E. changes with time,but angular momentum remains constant.
Statement -2 : No torque is acting on the rotating planet.So its angular momentum is constant.
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Solution
(b) Statement -1 is true, Statement-2 is true;Statement -2 is a correct explanation for Statement-1
Directions: Each question contains STATEMENT-1 and STATEMENT-2. Choose the correct answer(ONLY ONE option is correct ) from the following
Statement -1 :The escape speed does not depend on the direction in which the projectile is fired.
Statement -2 :Attaining the escape speed is easier if a projectile is fired in the direction the launch site is moving as the earth rotates about its axis.
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Solution
(c)Statement -1 is true, Statement-2 is true; Statement -2 is not a correct explanation for Statement-1
Directions: Each question contains STATEMENT-1 and STATEMENT-2. Choose the correct answer(ONLY ONE option is correct ) from the following
Statement -1 : If an object is projected from earth surface with escape velocity path of object will be parabola.
Statement -2 :When object is projected with velocity less than escape velocity from horizontal surface and greaterthan orbital velocity. Path of object will be ellipse.
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Solution
(c)Statement -1 is true, Statement-2 is true; Statement -2 is not a correct explanation for Statement-1
Directions: Each question contains STATEMENT-1 and STATEMENT-2. Choose the correct answer(ONLY ONE option is correct ) from the following
Statement -1 : Space rockets are usually launched in the equatorial line from west to east.
Statement -2 : The acceleration due to gravity is minimum at the equator.
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Solution
(c)Statement -1 is true, Statement-2 is true; Statement -2 is not a correct explanation for Statement-1
The speed of earth’s rotation about its axis is ω. Its speed is increased to x times to make the effective acceleration due to gravity equal to zero at the equator. Then x is
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Solution
g'= g – ω2 R cos2 λ(for equator λ = 0°)