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Chapter 6: Problem 790

A satellite of mass \(\mathrm{m}\) is orbiting close to the surface of the earth (Radius \(\mathrm{R}=6400 \mathrm{~km}\) ) has a K.E. \(\mathrm{K}\). The corresponding \(\mathrm{K} . \mathrm{E}\). of satellite to escape from the earth's gravitational field is (A) \(\mathrm{K}\) (B) \(2 \mathrm{~K}\) (C) \(\mathrm{mg} \mathrm{R}\) (D) \(\mathrm{m} \mathrm{K}\)

Short Answer

Expert verified
The satellite needs twice the kinetic energy as initially given to escape Earth's gravitational field. So, the correct answer is (B) \(2K\).

Step by step solution

01

Recall the formula for gravitational force and orbital speed

To solve this problem, we need to recall the formula for gravitational force, which is \[ F = G \frac{m M}{R^2} \] where F is the gravitational force, G is the gravitational constant (\(6.674 \times 10^{-11} \mathrm{Nm^2/kg^2}\)), m is the mass of the satellite, M is the mass of the Earth, and R is the radius (distance from Earth's center). For a satellite in orbit, the gravitational force equals the centripetal force required to keep the satellite in orbit, which is \[ F = \frac{m v^2}{R} \] where m is the mass of the satellite, v is the orbital speed, and R is the radius of its orbit.
02

Use the equations to find a relationship between K and orbital speed

By setting the gravitational force equal to the centripetal force, we derive the orbital speed, v, using the following equation: \[ G \frac{m M}{R^2} = \frac{m v^2}{R} \] Now, we find the kinetic energy(K) of the satellite in orbit, which is given by \[ K = \frac{1}{2} m v^2 \]
03

Determine the kinetic energy required to escape Earth's gravitational field

To escape Earth's gravitational field, the satellite needs to have enough kinetic energy to overcome the gravitational potential energy. The gravitational potential energy is: \[ U = G \frac{m M}{R} \] The total mechanical energy (sum of kinetic and potential energy) required to escape Earth's gravitational field is zero, so the required kinetic energy (K') is: \[ K' = -U \]
04

Establish the relationship between K and K' to find the correct answer

Now, we need to find the relationship between the given kinetic energy (K) and the kinetic energy needed to escape the gravitational field (K'). Substitute v in the K equation using the first equation in step 2: \[ K = \frac{1}{2}m \left( \frac{GM}{R} \right) \] Now, substitute K' in the U equation with the U given in step 3: \[ K' = G \frac{m M}{R} \] Comparing the above two equations, you can see that \[ K' = 2K \] So, the correct answer is (B) \(2K \). The satellite needs twice the kinetic energy as initially given to escape Earth's gravitational field.

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Most popular questions from this chapter

The escape velocity from the earth is about \(11 \mathrm{kms}^{-1}\). The escape velocity from a planet having twice the radius and the same mean density as the earth is \(=\ldots \ldots \ldots \ldots \mathrm{kms}^{-1}\). (A) 22 (B) 11 (C) \(5.5\) (D) \(15.5\)

The escape velocity for a body projected vertically upwards from the surface of earth is \(11 \mathrm{kms}^{-1}\). If the body is projected at an angle of \(45^{\circ}\) with the vertical, the escape velocity will be $\ldots \ldots \ldots \mathrm{kms}^{-1}$ (A) \((11 / \sqrt{2})\) (B) \(11 \sqrt{2}\) (C) 22 (D) 11

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