A body is projected up with a velocity equal to 34th of the escape velocity from the surface of the earth. The height it reaches from the centre of the Earth is?
farah tarannumBegginer
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To solve this problem, we need to consider the following:
1. Escape velocity: The escape velocity at Earth’s surface is approximately 11.2 km/s. So, the initial velocity of the body (3/4)th of escape velocity will be (3/4)*11.2 km/s = 8.4 km/s.
2. Potential energy: As the body moves upward, its kinetic energy converts into potential energy. At the maximum height, all the kinetic energy will be converted into potential energy.
3. Gravitational potential energy: The potential energy due to gravity at a distance R from the Earth’s center is given by: Potential Energy = -GMm/R, where G is the gravitational constant, M is the Earth’s mass, and m is the mass of the body.
4. Distance: We need to find the maximum height (R) reached by the body.
Solution:
1/2 mv^2 = GMm/R
where v is the initial velocity, m is the mass of the body, G is the gravitational constant, and M is the Earth’s mass.
1/2 (m)(8.4 km/s)^2 = 6.6743 x 10^-11 Nm^2/kg^2 (5.972 x 10^24 kg) (m)/R
R = (m)(8.4 km/s)^2 / (2 x 6.6743 x 10^-11 Nm^2/kg^2 x 5.972 x 10^24 kg)
Conclusion:
The body will reach a certain height based on its mass. However, we need to know the body’s mass to calculate the specific height it reaches from the Earth’s center.