Advertisements
Advertisements
Question
Determine the binding energy of satellite of mass 1000 kg revolving in a circular orbit around the Earth when it is close to the surface of Earth. Hence find kinetic energy and potential energy of the satellite. [Mass of Earth = 6 x 1024 kg, radius of Earth = 6400 km; gravitational constant G = 6.67 x 10-11 Nm2 /kg2 ]
Solution
Given:- m = 1000 kg, M = 6 x 1024 kg, R = 6400 km, G = 6.67 x 10-11 N m2/kg2
To find:-
i. Binding Energy (B.E.)
ii. Kinetic Energy (K.E.)
iii. Potential Energy (P.E.)
Formulae:- For satellite very close to earth,
i. B.E. =(1/2)*(GMm/R)
ii. K.E. = B.E.
iii. P.E. = -2K.E.
Calculation: From formula (i),
`B.E=(6.67xx10^-11xx6xx10^24xx1000)/(2xx6.4xx10^6)`
`B.E=(6.67xx6)/(12.8)xx10^10`
= antilog[log 6.67 + log 6 – log 12.8] x 1010
= antilog[0.8241 + 0.7782 – 1.1072] x 1010
= antilog[0.4951] * 1010
= 3.127 x 1010
∴ B.E. = 3.1265 x 1010 J
The binding energy of the satellite is 3.1265 x 1010 J.
From formula (ii),
K.E. = 3.1265 x 1010
∴ K.E. = 3.1265 x 1010 J
The kinetic energy of the satellite is 3.1265 x 1010 J.
From formula (iii),
P.E. = -2(3.1265 x 1010)
∴ P.E. = -6.2530 x 1010 J
The potential energy of the satellite is -6.2530 x 1010 J.
APPEARS IN
RELATED QUESTIONS
Find the total energy and binding energy of an artificial satellite of mass 800 kg orbiting at a height of 1800 km above the surface of the earth.
[G = 6.67 x 10-11 S.I. units, Radius of earth : R = 6400 km, Mass of earth : M = 6 x 1024 kg]
The escape velocity of a body from the surface of the earth is 11.2 km/s. If a satellite were to orbit close to the surface, what would be its critical velocity?
Define binding energy and obtain an expression for binding energy of a satellite revolving in a circular orbit round the earth.
