HSC Science (Electronics) 12th Board ExamMaharashtra State Board
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# 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. - HSC Science (Electronics) 12th Board Exam - Physics

ConceptBinding Energy and Escape Velocity of a Satellite

#### 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.

Is there an error in this question or solution?

#### APPEARS IN

2014-2015 (October) (with solutions)
Question 4.2 | 3.00 marks
Solution 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. Concept: Binding Energy and Escape Velocity of a Satellite.
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