5. Calculate the work done in moving a body of mass 1000 kg from a height
a height 3R above the surface of the Earth, where R = 6400 km is the radius of the
Earth.
Answers
The work done in moving the body is 2.615 x 10^0 J
Explanation:
Given data:
- Mass of body = 1000 Kg
- Height "h" = 3R
- Radius of earth "R" = 6400 Km
Solution:
r1 = R + h1 = R + 2R = 3R
r2 = R + h2 = R + 3R = 4R
W = GmM / 2r1 - GmM / 2r2 = Gm [ M / 3R - M / 4R]
W = GmM /2R (1/3 - 1/4 ) = GmM /2R ( 4 - 3/ 12)
W = GmM /24 R = 6.67 x 10^-11 x 6 x 10^24 x 1000 / 24 x 6.4 x 10^6
W = 6.67 x 6 x 10^10 / 24 x 6.4 = 2.615 x 10^0 J
Hence the work done in moving the body is 2.615 x 10^0 J
Also learn more
An artificial satellite revolves around the Earth at a distance of 3400 km. Calculate its orbital velocity and period of revolution. Radius of the Earth = 6400 km; g = 9.8 m/s² ?
https://brainly.in/question/195090
Answer:
I hope you will understand
Explanation:
We know, Mass of body(m) = 1000 kg, Mass of earth(M)= 6 × 10^24 kg , G = 6.67×10-11 N-m2 kg-2
work done to move a body = Change in Potential Energy (PE)
But, Change in (PE) = Final PE - Initial PE
Final PE = PE at height 3R , Initial PE = PE at height 2R
PE is given by equation = - GMm/R
Final PE = - GMm/3R , Initial PE = - GMm/2R
work done = - GMm/3R - (- GMm/2R) = - GMm/3R + GMm/2R
work done = (- 2GMm+ 3GMm) /6R = GMm/6R
work done = (6.67×10-11) x (6 × 10^24)x 1000 / (6x 6.4 x 10^6)
work done = 1.042 x 10^10 J