A particle of mass \( m \) is at a distance \( 3R \) from the center of Earth. Find the minimum kinetic energy of the particle to leave Earth's field, where \( R \) is the radius of Earth.
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The minimum kinetic energy required for a particle to escape Earth's gravitational field is equal to the magnitude of the gravitational potential energy at the given distance.
The work-energy theorem dictates that the minimum kinetic energy a particle needs to escape Earth's gravitational field equals the absolute value of the gravitational potential energy at its location.
The gravitational potential energy \( U \) at a distance \( r \) from Earth's center is defined as:
\[
U = -\frac{GMm}{r}
\]
where \( G \) is the gravitational constant, \( M \) is Earth's mass, \( m \) is the particle's mass, and \( r \) is the distance from Earth's center.
For this problem, the particle is situated at a distance of \( 3R \) from Earth's center, with \( R \) being Earth's radius. Consequently, the potential energy at this point is:
\[
U = -\frac{GMm}{3R}
\]
The minimum kinetic energy required for the particle to escape Earth's field is the energy needed to counteract this potential energy. Therefore, the minimum kinetic energy \( K \) is:
\[
K = \left| U \right| = \frac{GMm}{3R}
\]
Using the relationship \( g = \frac{GM}{R^2} \) (where \( g \) is the acceleration due to gravity at Earth's surface), we can express the kinetic energy as:
\[
K = \frac{gmR}{3}
\]
Thus, the minimum kinetic energy for the particle to escape Earth's field is \( \frac{2}{3} mgR \).
The correct answer is therefore (3) \( \frac{2}{3} mgR \).
