How to solve this? It’s a question on gravitation
Answers
Answer:
Explanation:
′=/(+ℎ)^2
in this case the g' = 0
I think you can replace value and solve after wards hope it helps
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× 10^6 m.
It is at a distance of 8 * 10^6 m from surface of earth
Hope it helps!