What is the value of orbital velocity when satellite just revolves around the earth?

When the satellite is very close to the earth, then r ~ R.

∴ the orbital velocity,


     straight v subscript straight o equals square root of GM over straight R end root equals square root of gR  

        equals space square root of 9.81 cross times 6.4 cross times 10 to the power of 6 end root space straight m divided by straight s 

         
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#6 {main}</pre> 

        space space equals 7.923 km divided by straight s
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Define orbital velocity. Derive the expression for it.

Orbital velocity is the velocity given to artificial satellite so that it may start revolving around the earth.

Orbital velocity is the velocity given to artificial satellite so tha

Expression for orbital velocity:
Suppose a satellite of mass m is revolving around the earth in a circular orbit of radius r, at a height h from the surface of the earth. Let M be the mass of the earth and R be radius of the earth.
∴ r = R + h
    To revolve the satellite, centripetal force of <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array)
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#6 {main}</pre> is required (where straight v subscript straight o is orbital velocity)  which is provided by gravitational force <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array)
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#6 {main}</pre> between the earth and the satellite.
∴               mv subscript straight o squared over straight r space equals space GMm over straight r squared
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#6 {main}</pre>

or               straight v subscript straight o space equals space square root of GM over straight r end root space equals space square root of fraction numerator GM over denominator straight R plus straight h end fraction end root
But GM=gR2 where g is the acceleration due to gravity.
straight v subscript straight o space equals space square root of gR squared over straight r end root space equals space square root of fraction numerator gR squared over denominator straight R plus straight h end fraction end root space equals space straight R square root of fraction numerator straight g over denominator straight R plus straight h end fraction end root
Let g' be the acceleration due to gravity in the orbit i.e. at a height h from the surface.
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#6 {main}</pre>
or            fraction numerator GM over denominator straight R plus straight h end fraction equals straight g apostrophe left parenthesis straight R plus straight h right parenthesis space equals space straight g apostrophe straight r
∴                space space space straight v subscript straight o space equals space square root of straight g apostrophe left parenthesis straight R plus straight h right parenthesis end root space equals space square root of straight g apostrophe straight r end root

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Define escape velocity. Derive an expression for escape velocity. Find the value of escape velocity.


Escape velocity is minimum velocity with which a body must be thrown upward so that it may just escape.
Expression for escape velocity:
Let a body of mass m be escaped from gravitational field of the earth. During the course of motion, let at any instant body be at a distance r from the centre of the earth.

Escape velocity is minimum velocity with which a body must be thrown
The gravitational force between body and the earth is,
                               straight F equals GMm over straight r squared dr
and work done to raise the body by distance dr is,
                             space space dW equals GMm over straight r squared dr
   Total work done W in raising the body from the surface of the earth to infinity is,
                             straight W equals integral subscript straight R superscript infinity GMm over straight r squared dr equals GMm open square brackets negative 1 over straight r close square brackets subscript straight R superscript infinity
                                 equals GMm open square brackets negative 1 over infinity plus 1 over straight R close square brackets equals GMm over straight R

If we throw the body upward with a velocity ve, then work done to raise the body from surface of the earth to infinity is done by kinetic energy,
                                 GMm over straight R equals 1 half mv subscript straight e squared
or                                straight v subscript straight e equals square root of fraction numerator 2 GM over denominator straight R end fraction end root space equals space square root of 2 gR end root
or                                 straight v subscript straight e equals square root of 2 gR end root
Substituting the values, g = 9.81m/sec and R = 6.4 x 106m, we have,
  straight v subscript straight e equals square root of 9.81 cross times 6.4 cross times 10 to the power of 6 end root space equals space 11200 space straight m divided by straight s
       equals 11.2 space km divided by straight s

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Show that the satellite revolving around the earth will depart forever if its speed is increased by 41.42%.

Let a satellite of mass m be revolving in circular orbit of radius 'r' around the earth of mass M.

The orbital velocity of satellite is given by, 

                    straight v subscript straight o equals square root of GM over straight r end root 

The velocity required by the satellite is obtained from the equation,

             1 half mv subscript straight e squared space equals space GMm over straight r 

rightwards double arrow                  straight v subscript straight e equals square root of fraction numerator 2 GM over denominator straight r end fraction end root 

To escape, the percentage increase in velocity is, 

fraction numerator straight v subscript straight e minus straight v subscript straight o over denominator straight v subscript straight o end fraction cross times 100 space equals space fraction numerator square root of begin display style fraction numerator 2 GM over denominator straight r end fraction end style end root space minus space square root of begin display style GM over straight r end style end root over denominator square root of begin display style GM over straight r end style end root end fraction cross times 100 

                   space space equals left parenthesis square root of 2 minus 1 right parenthesis cross times 100 space

space space equals space 41.42 percent sign
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Two bodies of masses m1 and m(m1 > m2) are to be projected out of the field of the earth.
(i) Which of the two must be thrown with greater velocity?
(ii) Which of the two must be thrown with greater momentum?
(iii) Which of the two must be thrown with greater kinetic energy?


(i) Escape velocity is independent of the mass of body projected and is same for all the bodies.

i.e., 11.2 km/s, therefore, both must be projected with equal velocity. 

(ii) Momentum of body of mass m1 is,
                      straight p subscript 1 equals straight m subscript 1 straight v subscript straight e 

Momentum of the body of mass straight m subscript 2 is, 

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#6 {main}</pre> 

Since    straight m subscript 1 greater than straight m subscript 2 

∴             space space space space space space space straight m subscript 1 straight v subscript straight e greater than straight m subscript 2 straight v subscript straight e 

That is, a heavy body must be thrown with greater momentum. 

(iii) Kinetic energy of mass straight m subscript 1 is, 

           space space space space space space straight K subscript 1 equals 1 half straight m subscript 1 straight v subscript straight e squared 

Kinetic energy of mass straight m subscript 2 is,

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#6 {main}</pre> 

Since m1 > m2, therefore K1 > K2.

That is, i.e., heavy body must be thrown with greater kinetic energy. 

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