L9 : Examples-Gravitational POtential energy - Gravitation, Physics, Class 11 Video Lecture - Class 12 | Best Video for Class 12

hello friends this video on gravitation
part 9 is brought to you by exam feel
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going ahead with path 9 kills all
certain problems to clarify the idea of
potential energy or gravitational
potential energy which we'll study till
now so let us look at problem number 1
it says to earth satellites a and B each
of mass M are to be launched into
circular orbits above Earth's surface at
altitudes 6400 kilometer and 1.9 jewel
to 10 to the power 4 kilometers
respectively the radius of the earth is
6 4 0 0 kilometers of their potential
energies and the ratio of their kinetic
imagism
which one has total greater total energy
so let us first consider the first part
the first part asks us to find out the
ratio of their potential energies so
what all are given here mass of a the
first satellite is M mass of the second
satellite bead is also abundant in now
they are launched in circular orbits
above Earth's surface at altitudes 6,400
kilometers and 1.9 2 into 10 to the
power 4 kilometers that means the height
for a is equal to 6 4 0 0 kilometers
and the height for B is equal to 1 point
9 2 into 10 to the power 4 kilometers
nowaday Diaz of the earth that is our e
is equal to 6 4 0 0 kilometers to the
ratio of the potential energies see when
I say this is the earth and this is a
satellite let us suppose so the
satellite is moving who would be moving
around the earth somewhat like this
right it says that they are launched
into circular orbits above Earth's
surface at altitudes 6400 kilometers
that means when I talk of H H would mean
this distance but when I calculate the
distance between Earth and the satellite
it would be the radius of the earth plus
the height of the satellite ok so we can
say that let us calculate the potential
energy first so potential energy is
equal to minus G capital M into small M
divided by R that is the total distance
R is the distance of the satellite from
the earth so what would be the distance
it would be the radius of the earth plus
the height of that particular satellite
this capital n is the mass of the earth
now let us calculate the potential
energy for a and then for B so for a the
potential energy would be equal to minus
g m-e into m divided by re plus h that
means 6 4 0 0 plus h a h a is again 6 4
0 0 so this will be minus g m-e into m /
1 - 8 0 0 now we will calculate the
potential energy falling
that would be potential energy for a is
equal to minus g m-e into m / r e is
again 6 4 0 0 + 1 point 9 2 into 10 to
the power 4 so this comes out doing
minus g m-e
into n divided by 6 4 0 0 + 1 9 - 0 0
the ratio of potential energies that is
the ratio of potential energies should
be equal to potential energy of a
divided by potential energy of being so
this would be equal to minus g m-e into
m / 1 - 8 0 0 into 6 4 0 0 + 1 9 2 0 0 /
- chain M into M so these two will get
cancelled so this value comes out to be
2 by 1 that is equal to 2 is to 1
therefore the ratio of the potential
energies is equal to 2 is to 1 so in
this case all you need to understand is
what would be the distance I mean this
thing should be clear that word when you
are calculating potential energy it is -
GMP in GM divided by R where R is the
distance between the satellite and the
earth and the question saying that the
at satellite is at a height of six four
zero zero kilometer above the surface of
the earth that means the total distance
will be the radius of the earth that is
this is the radius of the earth and this
is the height of
Saturday so the total distance of
satellite from Earth would be our e + H
so this is how we calculated the ratio
of the potential energy so till now we
solved the first part right we found the
ratio of the potential energy now let us
try to find the ratio of the kinetic
energies so for that what we will do
first of all you will calculate the
kinetic energy of the satellite a so
kinetic energy of satellite a would be
equal to G capital m into small M
divided by 2 into R so again what is
this this would be nothing but one two
eight zero zero right now similarly we
will calculate the kinetic energy of
satellite B let us denote that by Kb
so when this would become capital G into
M into small M divided by 2 into 1 point
9 2 into 10 to the power 4 plus 6 4 0 0
so the same rule applies here also here
you can see that this R go to the side
this is the radius of the earth plus the
distance of the satellite as I already
explained before that let us suppose if
this is the earth so this is the radius
of the earth right and let us suppose
this is your satellite so this distance
Plus this distance would be the total R
that means the total distance of the
satellite from the center of the earth
that means it is the radius of the earth
plus the height of the satellite from
the surface of the earth so we applied
the same rule for a as well as B final
now let us calculate the ratio so the
ratio would be K a by K B so now if you
divide this by this you would get G
Capital m in
small M divided by 2 into 1 2 8 0 0 into
2 into 1 point 9 2 into 10 to the power
4 + 6 4 0 0 divided by G into capital M
into small M so this three terms will
get cancelled this two will get
cancelled with this 2 so we find in that
this value comes out to be 2 by 1 so the
ratio is again 2 is to 1 so the ratio of
the kinetic energies of a and B is
nothing but 2 is to 1 now what does the
last part of the question asks it asks
which one has greater total energy let
us look at that now what would be the
total energy of a as per the expression
we know that the total energy would be
minus G into Capital m into small M
divided by 2 I where R is equal to 1 2 8
0 0 right this is the value of R ok now
similarly let us write what would be the
total energy of B so we apply the same
formula G minus G into capital M into
small M divided by 2 R but in this case
R would be 1 point 9 2 into 10 to the
power 4 plus 6 4 0 0 so this would be
the value of pi now if I ask you which
will have greater total energy see if
you see if you compare these two both
and have both have the same numerator
right so the one with whose denominator
is smaller would have more total energy
so you can see the value of I the
denominator is dependent on the value of
I so the one which has lesser value of R
will have greater energy so R is more in
which case the value of R is more for
this
right so the value of AD is more for a
that means the total energy for a would
be less they're poor which has greater
energy total energy of B is greater than
a fine okay
so now we'll go ahead and look further
on different targets thank you please
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