17) A rocket is fired vertically with a speed of 5 kms-1 from the eart

1.	17) A rocket is fired vertically with a speed of 5 kms-1 from the earth's surface. How far from theearth does the rocket go before returning tothe earth ? Mass of the earth 60 x 1024 kgmean radius of the earth6.4 x 106 m;G6.67 x 10-11 Nm2 kg2,itil eneed of rocket is v and at
Answer» Mass of the earth (Me) = 6×10²⁴ Kg Mean orbital radius of earth around the sun (r) = 6.4 × 10^6m P.E of the rocket at earth's surface (Uo) = -GMem/Re P.E of the rocket at height h from the earth's surface. Uh = -GMem/(Re + h) Increase in PE (∆U) = Uh-Uo = -GMem/(Re + h) - (-GMem)/Re = GMem[ 1/Re - 1/(Re + h)] = GMem× h/Re(Re + h) g = GMe/Re² GMe = gRe² ∆U = gRe² × m×h/Re(Re+h) = mgh/( 1 + h/Re) According to law of conservation of energy . Increase in P.E = KE of the rocket 1/2 × mv² = mgh/(1 + h/Re) v²( 1 + h/Re) = 2gh h = v²Re/( 2gRe - v²) Here, v = 5 km/s = 5000 m/s Re = 6.4 × 10^6 m g = 9.8 m/s² h = (5 × 10³)²× 6.4×10^6/{ (2×9.8×6.4×10^6) - (5×10³)²}= 1600 km Distance from the centre of the earth = h + Re = 6400 + 1600 = 8000 km = 8× 10^6m hit like if you find it useful

17) A rocket is fired vertically with a speed of 5 kms-1 from the earth's surface. How far from theearth does the rocket go before returning tothe earth ? Mass of the earth 60 x 1024 kgmean radius of the earth6.4 x 106 m;G6.67 x 10-11 Nm2 kg2,itil eneed of rocket is v and at

Answer»

Mass of the earth (Me) = 6×10²⁴ Kg

Mean orbital radius of earth around the sun (r) = 6.4 × 10^6m

P.E of the rocket at earth's surface (Uo) = -GMem/Re

P.E of the rocket at height h from the earth's surface.

Uh = -GMem/(Re + h)

Increase in PE (∆U) = Uh-Uo

= -GMem/(Re + h) - (-GMem)/Re

= GMem[ 1/Re - 1/(Re + h)]

= GMem× h/Re(Re + h)

g = GMe/Re²

GMe = gRe²

∆U = gRe² × m×h/Re(Re+h)

= mgh/( 1 + h/Re)

According to law of conservation of energy .

Increase in P.E = KE of the rocket

1/2 × mv² = mgh/(1 + h/Re)

v²( 1 + h/Re) = 2gh

h = v²Re/( 2gRe - v²)

Here, v = 5 km/s = 5000 m/s

Re = 6.4 × 10^6 m g = 9.8 m/s²

h = (5 × 10³)²× 6.4×10^6/{ (2×9.8×6.4×10^6) - (5×10³)²}= 1600 km

Distance from the centre of the earth = h + Re = 6400 + 1600 = 8000 km = 8× 10^6m

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