What is the mathematical relationship between the escape velocity (Ve) and the orbital velocity (Vo) of a satellite revolving very close to the surface of the Earth? MCQ with Answer and Explanation
What is the mathematical relationship between the escape velocity (Ve) and the orbital velocity (Vo) of a satellite revolving very close to the surface of the Earth?
A. Ve = sqrt(2) * Vo
B. Ve = Vo
C. Ve = 2 * Vo
D. Vo = sqrt(2) * Ve
Answer: Option A
Solution (By JKSSB Mock Tests)
The orbital velocity of a satellite very close to Earth is Vo = sqrt(gR). The escape velocity from Earth's surface is Ve = sqrt(2gR). Therefore, substituting Vo into the escape velocity equation yields Ve = sqrt(2) * Vo. A satellite requires a 41.4% increase in speed to escape orbit completely.
Explanation:
Magnetic dipole moment M = Current(I) * Area(A). Current is charge/time: I = e / T. Time period T = 2pir / v. So I = ev / (2pir). The area of the circular orbit is A = pir^2. Therefore, M = [ev / (2pir)] * [pir^2] = evr / 2. This is a standard derivation in modern physics.
Explanation:
Candela (cd) is SI base unit for luminous intensity. Lumen: luminous flux; Lux: illuminance; Watt: power. Memory aid: 'Candela = candle power; base unit for perceived light'. Unit knowledge frequently tested in competitive exams.
Explanation:
At maximum height, final velocity v = 0. Using v = u - gt (taking upward positive), 0 = u - gt ⇒ t = u/g. This is time of ascent. Total time of flight would be 2u/g. The equation derives from Newton's first equation of motion under constant acceleration g downward. Memory aid: Time to peak = initial velocity / gravitational acceleration. This fundamental result appears in projectile motion problems. Competitive exams often combine this with energy conservation or symmetry concepts for advanced questions.
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