The additional K.E. to be provided to a satellite of mass \(\mathrm{m}\) revolving around a planet of mass \(\mathrm{M}\), to transfer it from a circular orbit of radius \(\mathrm{R}_{1}\) to another radius \(\mathrm{R}_{2}\left(\mathrm{R}_{2}>\mathrm{R}_{1}\right)\) is (A) $\operatorname{GMm}\left[\left(1 / R_{1}^{2}\right)-\left(1 / R_{2}^{2}\right)\right]$ \(\operatorname{GMm}\left[\left(1 / R_{1}\right)-\left(1 / R_{2}\right)\right]\) (C) $2 \mathrm{GMm}\left[\left(1 / \mathrm{R}_{1}\right)-\left(1 / \mathrm{R}_{2}\right)\right]$ (D) $(1 / 2) \mathrm{GMm}\left[\left(1 / \mathrm{R}_{1}\right)-\left(1 / \mathrm{R}_{2}\right)\right]$

Short Answer

Expert verified
The additional kinetic energy needed to transfer a satellite from a circular orbit of radius $R_1$ to another radius $R_2 (R_2 > R_1)$ is: (D) $\frac{1}{2} GMm\left[\left(\frac{1}{R_1}\right)-\left(\frac{1}{R_2}\right)\right]$

Step by step solution

01

Find kinetic energy in the first (R1) orbit

To find the kinetic energy of the satellite in the orbit with radius R1, we can use the formula: \[T_1 = \frac{1}{2}mv_{1}^2\], where 'v1' is the initial orbital speed of the satellite in the R1 orbit. Since it is in a circular orbit, we can use the centripetal force, which is the gravitational force in this case: \[\frac{mv_{1}^{2}}{R_{1}} = \frac{GMm}{R_{1}^{2}}\].
02

Solve for 'v1'

Now, we need to solve for the initial orbital speed 'v1' using the centripetal force equation: \[v_{1}^2 = \frac{GM}{R_1}\].
03

Substitute 'v1' into the kinetic energy formula for the R1 orbit

We substitute the expression we got for v1 into the kinetic energy formula: \[T_1 = \frac{1}{2}m \cdot \frac{GM}{R_1}\].
04

Find kinetic energy in the second (R2) orbit

We repeat steps 1 to 3 for the R2 orbit. The kinetic energy of the satellite in the orbit with radius R2 is given by: \[T_2 = \frac{1}{2}mv_{2}^2\], So, \[\frac{mv_{2}^{2}}{R_{2}} = \frac{GMm}{R_{2}^{2}}\], \[v_{2}^2 = \frac{GM}{R_2}\], \[T_2 = \frac{1}{2}m \cdot \frac{GM}{R_2}\].
05

Calculate the difference in kinetic energy

To find the additional kinetic energy needed to transfer the satellite from the R1 orbit to the R2 orbit, we calculate the difference in kinetic energy: \[\Delta T = T_2 - T_1 = \frac{1}{2}m \cdot \frac{GM}{R_2} - \frac{1}{2}m \cdot \frac{GM}{R_1}\].
06

Simplify the expression

Now we simplify the expression for the difference in kinetic energy: \[\Delta T = \frac{1}{2} GMm\left(\frac{1}{R_2} - \frac{1}{R_1}\right)\]. Comparing the simplified expression with the choices given, we see that the additional kinetic energy needed to transfer the satellite from the R1 orbit to the R2 orbit is: (D) $(1 / 2) \mathrm{GMm}\left[\left(1 / \mathrm{R}_{1}\right)-\left(1 / \mathrm{R}_{2}\right)\right]$

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Most popular questions from this chapter

Which one of following statements regarding artificial satellite of earth is incorrect (A) The orbital velocity depends on the mass of the satellite (B) A minimum velocity of \(8 \mathrm{kms}^{-1}\) is required by a satellite to orbit quite close to the earth. (C) The period of revolution is large if the radius of its orbit is large (D) The height of geostationary satellite is about \(36000 \mathrm{~km}\) from earth

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