Question

A rocket is fired vertically with a speed of 5 km s–1 from the earth’s surface. How far from the earth does the rocket go before returning to the earth? Mass of the earth = 6.0 × 1024 kg; mean radius of the earth = 6.4 × 106 m; G= 6.67 × 10–11 N m2 kg–2. Question 8.17 Gravitation

Answer 1

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

Answer 2

Answer:

The answer is 8 into 10 to the power of 6