Question

An object is projected vertically upward from the surface of the earth on
upward from the surface of the earth of mass M with a velocity such
that the maximum heignt reached is eight times the radius R of the earth. Calculate:
(i) the initial speed of projection
(ii) the speed at half the maximum height.

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Answer 1

Explanation ⇒ Let the mass of the object projected be m.

(a). Now,  Initial potential energy =  -GMm/R

Final potential energy = -GMM/9R

Initial kinetic energy = 1/2mv²

Final kinetic energy = 0 [Since, v = 0 at top.]

Now, decrease in kinetic energy = increase in potential energy.

∴ 1/2 mv² = GMm(1/R - 1/9R)

∴ v²/2 = 8GM/9R

∴ v² = 16GM/9R

∴ v = 4/3√(GM/R)

∴ v = 4/3 × √gR

∴ v = 4√(gR)/3

Hence, the initial speed of projection = v = 4√(gR)/3

(b).

For speed at half separation. Distance from centre = R + 8R/2 = 5R

Using the same concept,

1/2 mv² - 1/2 mt² = GMm(1/R - 1/5R), where t is the speed at half journey.

∴ m × 16GM/9R -  mt² = 8GMm/5R

∴ 16GM/9R - t² = 8GM/5R

∴ t² = GM/R [16/9 - 8/5]

∴ t² = GM/R × [80 - 72]/45

∴ t² = 8GM/45R

∴ t = √(8GM/45R)

Hope it helps.

Answer 2

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