Question

How to solve this? It’s a question on gravitation

Answer 1

Answer:

Explanation:

′=/(+ℎ)^2

in this case the g' = 0

I think you can replace value and solve after wards hope it helps

Answer 2

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!