Physics, asked by Anonymous, 11 months ago

a satellite of mass Ms orbits the earth (with mass me) and radius Re with a velocity V and an altitude H . The gravity Fg and the centripetal force FC are given by

Fg=G Ms.Me/(Re+H)^2,Fc=Ms.v^2/Re+h. , G=const.
Find the equation for kinetic energy E
kin
of satellite
(b)Based on the kinetic energy ,how much liquid hydrogen (energy density 10^6J/litre ) is at least needed to bring a small 1 kg satellite in an orbit of 400 km (use literature to find Me,Re,G)​

Answers

Answered by abicharan27
20

Answer:

Let, a satellite of mass m goes around the earth in a circular orbit of radius r=R+R=2R. Here the centripetal force mv

2

/r necessary to keep the satellite in its circular orbit is provided by the gravitational force GMm/r

2

between the earth and the satellite.

Thus, mv

2

/r=GMm/r

2

or v=

r

GM

=

2R

GM

Acceleration at the earth's surface is g=

R

2

GM

So, v=

2R

gR

2

=

gR/2

Answered by mohamed1318046
6

Answer:

(a) The kinetic energy of the satellite = 1/2*G * ME*Ms / (RE+h)

(b) 33.12 Litre

Explanation:

(a) Let's assume RE + h = L. As Fc = Ms * v^2 / L and Fg = G * Ms*ME / L^2 and Kinetic energy = 1/2 M*v^2

As the centripetal force (in this situation) is the same as the gravitational force, we'll conclude the kinetic energy by saying that Fc = Fg which is Ms * v^2 / L = G * Ms*ME / L^2. This will end with Ms*v^2 = G * ME*Ms / L. Finally, by dividing both sides by 2 = 1/2 M*v^2 = 1/2*G * ME*Ms / L = 1/2*G * ME*Ms / (RE+h). So, the kinetic energy of the satellite = 1/2*G * ME*Ms / (RE+h)

(b) ME = 5.972 × 10^24 kg. RE = 6,371 km. G = 6.67 * 10^−11 N⋅m2/kg2. We will use both concepts of kinetic and potential energies. As kinetic energy = 1/2*G * ME*Ms / h and potential energy = - G * ME*Ms / h and Total energy = kinetic + potential. Total energy of the satellite at the surface of the earth (Eo) = 0 - G * ME*Ms / RE = - G * ME*Ms / RE. Total energy of the satellite when reached the orbit given (Ef) = 1/2*G * ME*Ms / (RE+h) - G * ME*Ms / (RE+h) = -1/2*G * ME*Ms / (RE+h).

So, to get the energy needed (E) for the satellite to have the ability to reach the given orbit, we'll subtract the final energy from the initial one. E = Ef - Eo = -1/2*G * ME*Ms / (RE+h) - (- G * ME*Ms / RE) = G * ME*Ms * ( 1/RE - 1 / 2(RE+h)) =  

(6.67 * 10^−11) * (5.972 × 10^24)*(1) * ( 1/(6371*10^3 - 1 / 2(6771*10^3)) = 3.3 * 10^7 Joules. As the energy density = 10^6 L/Litre for the liquid hydrogen, the total volume of liquid hydrogen needed = 3.3 * 10^7 / 10^6 = 33.12 Litre

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