The total energy of satellite is –
A) Always positive
B) Always negative
C) Always zero
D) +ve or –ve depending upon radius of orbit
Answer
558.3k+ views
Hint: The energy required to revolve around the earth by a satellite is known as orbital energy. As satellites revolve around the earth so it requires some energy known as kinetic energy to keep hold there and also gravitational potential energy as satellite is present in the gravitational field of earth.
Complete step-by-step answer:
There are two energies of a satellite that holds it around the earth:
a) Kinetic energy –
By considering the earth as reference and
r = radius of earth,
M = mass of satellite,
m = mass of earth,
v = velocity
We can find gravitational force (${{F} _ {g}} $) and centrifugal force (${{F} _{c}} $)–
${{F} _ {g}} =\dfrac {mM} {{{r} ^ {2}}} $ and,
${{F} _{c}} =m {{v} ^ {2r}} $
As these above two forces are equal so,
${{F} _ {g}} = {{F} _{c}} $
$\dfrac {mM} {{{r} ^ {2}}} =m {{v} ^ {2}} r$
${{v} ^ {2}} =\dfrac {GM} {r} $ ……………………………. (1)
Putting (1) in formula of kinetic energy we get:
$K.E.=\dfrac{1}{2}m{{v}^{2}}$
$K.E.=\dfrac {1} {2} m (\dfrac{GM}{r})$
$\therefore K.E.=\dfrac{GmM}{2r}$ ……………………………….. (2)
b) Gravitational potential energy –
Let us consider,
$r_1$= distance of mass m,
$r_2$ = distance of earth from center,
$M$ = mass of satellite,
$m$ = mass of earth,
$F$= force,
$\therefore \,\Delta U=\int\limits_{{{r}_{1}}}^{{{r}_{2}}}{F.dr}$
$=GmM\int\limits_ {{{r} _ {1}}} ^ {{{r} _ {2}}} {dr} $
$=GmM(\dfrac{1}{{{r}_{1}}}-\dfrac{1}{{{r}_{2}}})$
So, for the equation of U we already know that, $\Delta U= {{U} _ {2}}-{{U} _ {1}} $
$\therefore U=-\dfrac {GmM} {r} $ …………………………………… (3)
From equation (2) $\Rightarrow $ (3):
We write the total energy of a satellite as the sum of its kinetic energy and the gravitational energy it have –
$\therefore Total\,energy=\,K.E.+G.P.E$
Where, K.E. = kinetic energy,
G.P.E. = gravitational potential energy.
$Total\, energy=\dfrac {GmM} {2r} +\left (\dfrac{-GmM}{r} \right)$
$Total\, energy=-\dfrac {GMm} {2r} $
Hence we can see that the total energy of the satellite is always negative.
So, the correct answer is “Option B”.
Note: Satellite is launched by rocket in the orbit of the earth and once it is successfully launched it revolves around the earth to perform its task therefore, to stay in the earth’s orbit satellite needs some form of energies. Every satellite has its own purpose of launch and our country, India is one of the leading countries in successfully launching satellites.
Complete step-by-step answer:
There are two energies of a satellite that holds it around the earth:
a) Kinetic energy –
By considering the earth as reference and
r = radius of earth,
M = mass of satellite,
m = mass of earth,
v = velocity
We can find gravitational force (${{F} _ {g}} $) and centrifugal force (${{F} _{c}} $)–
${{F} _ {g}} =\dfrac {mM} {{{r} ^ {2}}} $ and,
${{F} _{c}} =m {{v} ^ {2r}} $
As these above two forces are equal so,
${{F} _ {g}} = {{F} _{c}} $
$\dfrac {mM} {{{r} ^ {2}}} =m {{v} ^ {2}} r$
${{v} ^ {2}} =\dfrac {GM} {r} $ ……………………………. (1)
Putting (1) in formula of kinetic energy we get:
$K.E.=\dfrac{1}{2}m{{v}^{2}}$
$K.E.=\dfrac {1} {2} m (\dfrac{GM}{r})$
$\therefore K.E.=\dfrac{GmM}{2r}$ ……………………………….. (2)
b) Gravitational potential energy –
Let us consider,
$r_1$= distance of mass m,
$r_2$ = distance of earth from center,
$M$ = mass of satellite,
$m$ = mass of earth,
$F$= force,
$\therefore \,\Delta U=\int\limits_{{{r}_{1}}}^{{{r}_{2}}}{F.dr}$
$=GmM\int\limits_ {{{r} _ {1}}} ^ {{{r} _ {2}}} {dr} $
$=GmM(\dfrac{1}{{{r}_{1}}}-\dfrac{1}{{{r}_{2}}})$
So, for the equation of U we already know that, $\Delta U= {{U} _ {2}}-{{U} _ {1}} $
$\therefore U=-\dfrac {GmM} {r} $ …………………………………… (3)
From equation (2) $\Rightarrow $ (3):
We write the total energy of a satellite as the sum of its kinetic energy and the gravitational energy it have –
$\therefore Total\,energy=\,K.E.+G.P.E$
Where, K.E. = kinetic energy,
G.P.E. = gravitational potential energy.
$Total\, energy=\dfrac {GmM} {2r} +\left (\dfrac{-GmM}{r} \right)$
$Total\, energy=-\dfrac {GMm} {2r} $
Hence we can see that the total energy of the satellite is always negative.
So, the correct answer is “Option B”.
Note: Satellite is launched by rocket in the orbit of the earth and once it is successfully launched it revolves around the earth to perform its task therefore, to stay in the earth’s orbit satellite needs some form of energies. Every satellite has its own purpose of launch and our country, India is one of the leading countries in successfully launching satellites.
Recently Updated Pages
Master Class 12 Business Studies: Engaging Questions & Answers for Success

Master Class 12 Biology: Engaging Questions & Answers for Success

Master Class 12 Chemistry: Engaging Questions & Answers for Success

Class 12 Question and Answer - Your Ultimate Solutions Guide

Master Class 11 Social Science: Engaging Questions & Answers for Success

Master Class 11 English: Engaging Questions & Answers for Success

Trending doubts
Which is more stable and why class 12 chemistry CBSE

Which are the Top 10 Largest Countries of the World?

Draw a labelled sketch of the human eye class 12 physics CBSE

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

What are the major means of transport Explain each class 12 social science CBSE

Sulphuric acid is known as the king of acids State class 12 chemistry CBSE

