Question

calculate the kinetic energy ,potential energy, total energy and binding energy of an artificial satellite of mass 2000 kg orbiting at a height of 3600 km above the surface of the Earth. Given :- G=6.67×10^-11Nm^2/kg^2 R=6400 KM M=6×10^24 kg (ans:KE=40 09×10^9, PE=-80.09×10^9, TE=40.09×10^9, BE=40.09×10^9j)​

Answer 1

Answer:

P.E= -GMm/(h+R)

= -(6.67×10^-11×6×10^24×2000)÷(3600+6400)×1000

=- 8.009×10^10

=-80.09×10^9J

p.s R and h are converted to meters✌️

B.E = GMm÷2R

=( 6.67×10-11×6×10^24×2000)÷ 2×6400

=40.09×10^9J

K.E = B.E

=40.09×10^9J

T.E = -B.E

= -40.09×10^9J

Answer 2

The kinetic energy of the satellite is $40.02\times10^{9}\ J$

The potential energy of the satellite is $-80.02\times10^{9}\ J$

The total energy of the satellite is $40.0\times10^{9}\ J$

The binding energy of the satellite is $40.02\times10^{9}\ J$

Explanation:

Given that,

Mass of satellite = 2000 kg

Height = 3600 km

We need to calculate the kinetic energy of satellite

Using formula of kinetic energy

$K.E=\dfrac{GMm}{2r}$

Here, r = R+h

Put the value into the formula

$K.E=\dfrac{6.67\times10^{-11}\times6\times10^{24}\times2000}{2(3600+6400)\times10^{3}}$

$K.E=40.02\times10^{9}\ J$

We need to calculate the potential energy of satellite

Using formula of potential energy

$P.E=-\dfrac{GMm}{r}$

Put the value into the formula

$P.E=\dfrac{6.67\times10^{-11}\times6\times10^{24}\times2000}{(3600+6400)\times10^{3}}$

$P.E=-80.02\times10^{9}\ J$

We need to calculate the total energy of satellite

Using formula of total energy

$T.E=-P.E+K.E$

Put the value into the formula

$T.E=-80.02\times10^{9}+40.02\times10^{9}$

$T.E=-40.00\times10^{9}\ J$

Here, Negative sign show that satellite bounded with gravitational force

We need to calculate the binding energy of satellite

Using formula of binding energy

$B.E=\text{energy of satellite at infinity distance - energy of satellite in earth orbit}$

$B.E=0-\dfrac{GMm}{r}$

Put the value into the formula

$B.E=0-(-40.02\times10^{9})$

$B.E=40.02\times10^{9}\ J$

Hence, The kinetic energy of the satellite is $40.02\times10^{9}\ J$

The potential energy of the satellite is $-80.02\times10^{9}\ J$

The total energy of the satellite is $40.00\times10^{9}\ J$

The binding energy of the satellite is $40.02\times10^{9}\ J$

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Topic : Energy

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