An object going vertical upward form the surface of the earth , having an initial velocity equal to the .......... , escapes the gravitational force of the earth.
Answers
Answer:
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EXPLANATIONWe know that the escape velocity of Earth = Ve = √[2gR]
g = gravity due to Earth
1/2 m (Ve)² - G Mm/R = 0 for the object to escape from Earth.
(or, KE + PE = 0)
When the body is projected from Earth's surface:
Given u = initial velocity = Ve /2 = √(gR/2)
Initial KE = 1/2 m u² = m g R/4 = G M m /(4R)
Initial PE = - GMm/R
Total initial energy = - 3 GMm /(4R)
When the body reaches a height h above surface of Earth and stops:
v = 0. Final KE = 0.
Final PE = - G M m/ (R+h)
From conservation of Energy:
- GMm/(R+h) = - 3Gm/ (4R)
3 (R+h) = 4 R
h = R/3
Answer is R/3.
ESCAPE VELOCITY (11.19 km/s)
An object going vertical upward from the surface of the earth, having an initial velocity equal to the ESCAPE VELOCITY (11.19 km/s), escapes the gravitational force of the earth
Explanation:
escape velocity is the minimum speed needed for an object to escape from the gravitational influence of a massive body like earth .
Escape velocity Ve is a function of the mass of the body and distance to the center of mass of the body.
Ve= √(2GM/R)
G is the universal gravitational constant (G ≈ 6.67×10⁻¹¹ m³/kgs²),
M the mass of the body to be escaped from ( mass of earth in case of earth)
6 × 10²⁴ kg
R the distance from the center of mass of the body to the object ( Radius of earth on surface of earth ) = 6400 km
Ve= 11.19 km/s
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