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Two satellites A and B of equal mass move in the equatorial pl

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Two satellites $A$ and $B$ of equal mass move in the equatorial plane of the earth, close to the earth's surface. Satellite $A$ moves in the same direction as that of the rotation of the earth while satellite $B$ moves in the opposite direction. Calculate the ratio of the kinetic energy of $B$ to that of $A$ in the reference frame fixed to the earth. $(g = 9.8 m / s^{2}$ and radius of the earth $= 6.37 \times 1 0^{6} m)$

Name: Video solution 1: Two satellites A and B of equal mass move in the equatorial plane of the earth, close to the earth's surface. Satellite A moves in the same direction as that of the rotation of the earth while satellite B moves in the opposite direction. Calculate the ratio of the kinetic energy of B to that of A in the reference frame fixed to the earth. (g=9.8 m/s2 and radius of the earth =6.37×106 m)
Uploaded: 2022-04-24T18:08:03.000Z
Duration: 2 min

Views: 5,282 students

Updated on: Apr 24, 2022

Book:

Physics Galaxy Mechanics Vol 1 (Ashish Arora)

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Two satellites

A

and

B

of equal mass move in the equatorial plane of the earth, close to the earth's surface. Satellite

A

moves in the same direction as that of the rotation of the earth while satellite

B

moves in the opposite direction. Calculate the ratio of the kinetic energy of

B

to that of

A

in the reference frame fixed to the earth.

(g = 9.8 m / s^{2}

and radius of the earth

= 6.37 \times 1 0^{6} m)

6 mins ago

Discuss this question LIVE

6 mins ago

Text solutionVerified

Let

ω_{A}

and

ω_{B}

be the absolute angular speeds of

A

and

B

. Since they are in the same orbit, their time periods must be the same, i.e.,

ω_{A} = ω_{B}

. Considering the dynamics of circular motion in the cases,

m ω_{A}^{2} R = \frac{GMm}{R ^{2}} \Rightarrow ω_{A} = \frac{g}{R} (∵ GM = gR^{2})

Similarly,

ω_{B} = \frac{g}{R}

And

ω_{A} = ω_{B} = \frac{9.8}{6.37 \times 1 0 ^{6}} = 124 \times 1 0^{- 5} rad s^{- 1}

Now

ω_{A E} = ω_{A} - ω_{E} = 124 \times 1 0^{- 5} - 7.3 \times 1 0^{- 5}

= 116.7 \times 1 0^{- 5} rad s^{- 1}

ω_{BE}

(velocity of

B

relative to

E)

= ω_{B} - (- ω_{E}) = ω_{B} + ω_{E} = 131.1 \times 1 0^{- 5} rad s^{- 1}

Therefore

\frac{K _{B}}{K _{A}} = \frac{\frac{1}{2} m ω _{BE}^{2} r ^{2}}{\frac{1}{2} m ω _{AE}^{2} r ^{2}} = \frac{131. 3 ^{2}}{116. 7 ^{2}} = 1.27

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Question Text	Two satellites $A$ and $B$ of equal mass move in the equatorial plane of the earth, close to the earth's surface. Satellite $A$ moves in the same direction as that of the rotation of the earth while satellite $B$ moves in the opposite direction. Calculate the ratio of the kinetic energy of $B$ to that of $A$ in the reference frame fixed to the earth. $(g = 9.8 m / s^{2}$ and radius of the earth $= 6.37 \times 1 0^{6} m)$
Updated On	Apr 24, 2022
Topic	Gravitation
Subject	Physics
Class	Class 11
Answer Type	Text solution:1 Video solution: 1
Upvotes	182
Avg. Video Duration	2 min

Text solutionVerified

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