Physics, asked by vsanskar7279, 11 months ago

Based on the kinetic energy how much liquid hydrogen is at least needed to bring a small 1kg satellite in an orbit of 400km.

Answers

Answered by Rajvardhan6238
0

Explanation:

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Answered by TejasMishra261
1

Answer:

)/(2V)=2x105

(J/m3

)

The value of Dk is thus seen to be proportional to the product of the material density

ρ=m/V and the square of the speed. For a near earth satellite of mass density of 2000

kg/m3 moving in orbit at v=sqrt(gR)=7.9x103

m/s , the kinetic energy density becomes

Dk=6.24x1010J/m3

. This is some 300,000 times more than that of the moving spear. One

can also talk about the kinetic energy density of a high speed rifle bullet made of

lead(mass density ρ=11.34·103

kg/m3

)moving at v=500m/s. Its energy density is Dk=

1.42x109

J/m3

. Its energy density is thus 44 times less than the satellite but some 7100

times larger than the thrown spear.

The question now arises how are the kinetic energy densities of the bullet and the satellite

so much larger than that of the thrown spear? The answer clearly is that the large kinetic

energy densities have come about by conversion of chemical energy derived from the

burning of rocket fuel and from the combustion of gun powder. It is also obvious from

the bullet case that relatively small amounts of explosive powder contain a great deal of

energy. Let us explore how much chemical energy is typically available in a combustion

process. Take the case of burning in a controlled manner an oxygen-hydrogen

stoichiometric gas mixture originally in a cryogenic liquid state. During combustion

water is produced and 572kJ/mole of heat and kinetic energy are released. A kilogram of

water contains 55.5moles, so that we can say that the energy density stored in the liquid

oxygen-hydrogen mixture is-

Dc=55.5x572x106

=3.17x1010(J/m3

)

This number is quite high compared to most kinetic energy densities and shows why it

takes relatively little volume of stored chemical energy in a bomb to produce very high

velocity fragments. A typical roadside bomb in Afghanistan probably contains some 30kg

of high explosive inside a metal casing of comparable mass. We can estimate that the explosion of such a bomb will release about 108

J of energy which will produce a deadly

spherical shock wave accompanied by flying casing fragments initially moving at

supersonic speeds.

The chemical energy stored in gasoline is about 45 megajoules per kilogram and its

density is about 720kg/m3

. Thus its energy density is Dc=3.24x1010 J/m3 or about the

same as that for a liquid oxygen-hydrogen mixture. It should be remembered that

stoichiometric gaseous mixture of O2 and H2 will have an energy density about a

thousand times smaller. A fact which should be kept in mind by individuals proposing the

use of uncompressed gaseous hydrogen as a means of automobile propulsion.

Energy stored in batteries is quite small compared to that of gasoline. Typically a leadacis battery stores about 146 kJ/kg, so that its energy density is –

Dc=146,000x5000=7.3x108

(J/m3

)

This means its energy density is some 44 times smaller than that for gasoline. Again one

sees that batteries, although great for short trips in golf carts, are not able to effectively

compete with gasoline as an energy source for longer distance transportation because of

their need for frequent recharging.

Next we go to the energy density stored in nuclear material. In fission a uranium U235 is

split my a neutron into two smaller fragments (say barium and krypton) plus the release

of three neutrons capable of producing a chain reaction. In one of these fission processes

about 200Mev=3.2x10 -11J of energy is released in the form of kinetic energy and heat.

This means that if one cubic meter of U235 containing about 1/(2.7x10 -29)=

3.7x10 28 atoms were to completely fission, the energy released would be-

Dn=(3.2x10-11)(3.7x1028)=1.18x1018 (J/m2

)

This is a truly large energy density exceeding the best chemical storage by some seven

orders of magnitude. The mass converted to energy in this cubic meter of U235 would,

according to Einstein’s formula, be m=E/c2

=1.18x1018/(3x108

)=13.1kg or about

13.1/(19900)=0.06% of the original mass. In an actual atomic bomb the energy release

per volume of fissioning uranium is less than these numbers by a couple orders of

magnitude but still very large. The Hiroshima bomb “Little Boy” contained 64kg of U235

but only about 1kg of that amount actually underwent fissioning.

Next we come to thermonuclear fusion. Here one has two isotopes of hydrogen at very

high temperature and pressure fusing into a helium atom plus the release of an energetic

neutron. The easiest way to achieve fusion involves the reaction-

D(deuterium)+T(tritium)→He(helium at 3.5Mev)+n(neutron at 14.1 Mev)

Explanation:

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