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KCET 2025 Physics Paper

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Question : 8 of 60

Marks: +1, -0

The total energy of a satellite in a circular orbit at a distance (

R+h

) from the centre of the Earth varies as [

R

is the radius of the Earth and

h

is the height of the oribit from Earth's surface]

$-\;\frac{1}{R+h}$
$\;\frac{1}{(R+h)^2}$
$-\;\frac{1}{(R+h)^2}$
$\;\frac{1}{R+h}$

Solution:

The total (mechanical) energy E of a satellite in a circular orbit of radius

r=R+h

is the sum of its kinetic and potential energies. You can show that

Gravitational potential energy:

U=-\;\frac{G M m}{r}

Kinetic energy (for a circular orbit):

K=\;\frac{1}{2} \;\frac{G M m}{r}

Total energy:

E=K+U=\;\frac{1}{2} \;\frac{G M m}{r}-\;\frac{G M m}{r}=-\;\frac{1}{2} \;\frac{G M m}{r}

Since

G, M

and

m

are constants,

E

varies as

E \propto -\;\frac{1}{r}=-\;\frac{1}{R+h}

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