Chapter 14, Problem 14A

a. What is the minimum speed for circling Earth in close orbit?

b. What is the maximum speed in an orbit that comes close to Earth at one point?

c. What happens to a satellite traveling faster than the maximum speed described in part (b)?

To determine

The minimum speed for circling Earth in close orbit.

Answer to Problem 14A

  $7.8km/s$

Explanation of Solution

Given:

For a satellite to maneuver in low Earth orbit while not falling down, minimum distance required from the Earth’s surface is $200km/s$ .

Formula used:

Centripetal Force $Fc=\frac{ms{v}^2}{r}$

Gravitational Force $Fg=\frac{G{m}_e{m}_s}{r^2}$

Where, $m_s$ is mass of satellite, $$ is mass of Earth, $v$ is speed of satellite, $r$ is distance from the center of Earth to the satellite, $G$ is Gravitational constant.

Calculation:

Force of gravity should balance the Centripetal Force.

Thus,

  $\frac{G{m}_e{m}_s}{r^2}=\frac{ms{v}^2}{r}$

Simplifying the equation would result in the speed of satellite revolving at low Earth orbit.

  $v=\sqrt{\frac{G{m}_e}{r}}$

Inserting the values of constants, speed of satellite would be

  $v=\sqrt{\frac{(6.67\times {10}^{-11}N.m^{2}k{g}^{-2})(5.98\times {10}^{24}kg)}{(6.38\times {10}^{6}m)+(2\times {10}^{5}m)}}=7.78km/s$

Conclusion:

Thus, minimum speed for a satellite to revolve in close orbit of Earth is 7.78km/s.

To determine

The maximum speed in an orbit that comes close to Earth at one point.

Answer to Problem 14A

  $11.2km/s$

Explanation of Solution

Given:

According to Energy conservation principle, total energy of an isolated system remains constant, neither it can be created nor can be destroyed. Total energy is equal to the sum of Potential energy $U_g=mgh$ and Kinetic energy $K=\frac{1}{2}m{v}^2$ of an object.

Calculation:

For an object to come closer to orbit or minimum speed required to escape Earth’s gravity is known as Escape velocity $v_e$ .

At this point, total energy of the object initial is equal to the final velocity after escaping the gravity at infinite distance $U_{gf}=0$ , and at that point final velocity is zero $K_f=0$ .

Thus,

  ${(K+U_g)}_i={(K+U_g)}_f$

  $\frac{1}{2}m{v}_e{}^2+\frac{-GMm}{r^2}=0+0$

  $v_e=\sqrt{\frac{2GM}{r}}$

Inserting the values of Gravitational constant, mass of Earth and distance from center of Earth to the object.

  $v_e=\sqrt{\frac{2\times (6.67\times {10}^{-11}N.m^{2}k{g}^{-2})(5.98\times {10}^{24}kg)}{(6.38\times {10}^{6}m)+(2\times {10}^{5}m)}}=11.2km/s$

Conclusion:

Thus, maximum speed in an orbit that comes close to Earth at one point or escape velocity is $11.2km/s$ .

To determine

To explain: The condition to a satellite traveling faster than the maximum speed .

Answer to Problem 14A

It will escape from Earth’s gravity.

Explanation of Solution

Introduction:

Escape velocity is the minimum speed of the satellite which without propulsion can escape Earth’s gravity.

For every celestial body like planet, minimum speed is required to escape from the gravitational field of that celestial body. This speed is said to be as Escape velocity. After reaching this speed, object escape out of gravity and it can travel to infinity without any propulsion or any resistance force because it will keep moving with the achieved speed.

Conclusion:

Thus, once a satellite travels faster than the maximum speed then it will get away from the gravitational field of the planet and can move to any said distance without any work done on the system.