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A satellite is moving with a constant speed v in circular orbi

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Solving time: 3 mins

A satellite is moving with a constant speed $v$ in circular orbit around the earth. An object of mass ' $m$ ' is ejected from the satellite such that it just escapes from the gravitational pull of the earth. At the time of ejection, the kinetic energy of the object is
(a) $\frac{3}{2} m v^{2}$
(b) $2 m v^{2}$
(c) $m v^{2}$
(d) $\frac{1}{2} m v^{2}$

Views: 6,252 students

Updated on: Feb 29, 2024

Book:

Arihant Physics JEE Main Chapterwise Solutions (2019-2002) (Arihant)

Exam:

JEE Mains 2019

Found 3 tutors discussing this question

Sebastian Discussed

A satellite is moving with a constant speed

v

in circular orbit around the earth. An object of mass '

m

' is ejected from the satellite such that it just escapes from the gravitational pull of the earth. At the time of ejection, the kinetic energy of the object is
(a)

\frac{3}{2} m v^{2}

(b)

2 m v^{2}

(c)

m v^{2}

(d)

\frac{1}{2} m v^{2}

12 mins ago

Discuss this question LIVE

12 mins ago

Text solutionVerified

Exp. (c)
Kinetic energy must be given to an object so that it just escapes to infinity is as shown in the figure below
Solution Image

Applying energy-conservation for mass

m

, we get
Initial energy of

m

at satellite

=

Final energy of

m

at infinity.

\Rightarrow U_{i} + K_{i} = U_{f} + K_{f} \dots (i)

Here,

U_{i} =

initial PE of

m

= \frac{- G \cdot M m}{r} = - (v^{2}) m

As,

v = \frac{GM}{r} =

orbital speed of the satellite.

K_{i} =

initial kinetic energy of

m

U_{f} =

final potential energy of

m

which is zero and

K_{f} =

final kinetic energy of

m

which is also zero.
So, from Eq. (i), we have

- v^{2} m + K_{i} = 0 + 0

K_{i} = m v^{2}

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Question Text	A satellite is moving with a constant speed $v$ in circular orbit around the earth. An object of mass ' $m$ ' is ejected from the satellite such that it just escapes from the gravitational pull of the earth. At the time of ejection, the kinetic energy of the object is (a) $\frac{3}{2} m v^{2}$ (b) $2 m v^{2}$ (c) $m v^{2}$ (d) $\frac{1}{2} m v^{2}$
Updated On	Feb 29, 2024
Topic	Gravitation
Subject	Physics
Class	Class 11
Answer Type	Text solution:1 Video solution: 5
Upvotes	580
Avg. Video Duration	5 min

Text solutionVerified

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