Chapter 14, Problem 22A

The positions of a satellite in elliptical orbit are indicated.

Rank these quantities from greatest to least.

a. gravitational force

b. speed

c. momentum

d. KE

e. PE

f. total energy $\left(KE+PE\right)$

g. acceleration

To determine

To rank: Gravitational force at different points from greatest to least.

Answer to Problem 22A

Gravitational force, from greatest to least is, A>B>C>D

Explanation of Solution

Given:

The position of satellite in elliptical orbit are shown below.

  

Formula used:

Gravitational force of attraction between two massive bodies of mass is $m_1$ and $m_2$ is,

  $F=G\frac{m_1{m}_2}{r^2}$

Where,

  $m_1$ = mass of the 1^st body.

  $m_2$ = mass of the 2^nd body.

G = Gravitational constant = $6.67\times {10}^{-11}\frac{\text{N}\cdot {\text{m}}^{\text{2}}}{{\text{kg}}^{\text{2}}}$

r = distance between to body.

Calculation:

From above equation, gravitational force is inversely proportional to the square of the distance between two body.

From figure, distance of the satellite from the earth is given at different position is,

(Position A) < (Position B) < (Position C) < (Position D)

Since less be the distance, large will be the force of gravity

  $\therefore F\propto \frac{1}{r^2}$

Conclusion:

Therefore, gravitational force, from greatest to least is, A > B > C > D.

To determine

To rank: Speed of satellite, from greatest to least.

Answer to Problem 22A

Speed of satellite, from greatest to least is, A > B > C > D

Explanation of Solution

Given:

The position of satellite in elliptical orbit are shown below.

  

Formula:

The speed of the satellite in elliptical orbit is,

  $\therefore v=\sqrt{GM\left(\frac{2}{r}-\frac{1}{a}\right)}$

Where,

G = Gravitational constant = 6.67×10-11 N.m²/kg²

M = Mass of the satellite

r = The distance of the satellite from the earth

a = Semi-major axis

Calculation:

From given formula,

  $\therefore v^2\propto \frac{1}{r}$

From the given formula, distance of the satellite increases as the satellite near to the earth. So, the speed is increases from position A to position D

Conclusion:

Therefore, speed of satellite, from greatest to least is, A>B>C>D

To determine

To rank: Momentum of satellite, from greatest to least.

Answer to Problem 22A

Momentum of satellite, from greatest to least is, A>B>C>D

Explanation of Solution

Given:

The position of satellite in elliptical orbit are shown below.

  

Formula:

The momentum of the satellite in elliptical orbit is,

  $\therefore P=Mv$

Where,

M = Mass of the satellite

v = Speed of the satellite.

P = momentum of the satellite.

Calculation:

Since from the given formula, momentum of the satellite is directly proportional to the velocity of the satellite.

  $\therefore P\propto v$

As from the given formula, speed of the satellite increases as the satellite near to the earth. So, the momentum is increases from D position A

Conclusion:

Therefore, momentum of satellite, from greatest to least is, A>B>C>D

To determine

To rank: Kinetic energy of satellite, from greatest to least.

Answer to Problem 22A

Kinetic energy of satellite, from greatest to least is, A>B>C>D

Explanation of Solution

Given:

The position of satellite in elliptical orbit are shown below.

  

Formula:

The kinetic energy of the satellite in elliptical orbit is,

  $\therefore KE=\frac{1}{2}M{v}^2$

Where,

KE = kinetic energy of satellite

M = Mass of the satellite

v = Speed of the satellite.

Calculation:

Since from the given formula, kinetic energy of the satellite is directly proportional to the velocity of the satellite.

  $\therefore KE\propto v^2$

From the given formula, speed of the satellite increases as the satellite near to the earth. So, the kinetic energy is increases from position D to position A

Conclusion:

Therefore, momentum of satellite, from greatest to least is, A>B>C>D

To determine

To rank: Potential energy of satellite, from greatest to least.

Answer to Problem 22A

Potential energy of satellite, from greatest to least is, D>C>B>A

Explanation of Solution

Given:

The position of satellite in elliptical orbit are shown below.

  

Formula:

The potential energy of the satellite in elliptical orbit is,

  $\therefore PE=-\frac{GMm}{r+R}$

Where,

G = Gravitational constant= 6.67×10-11 N.m²/kg²

M = Mass of the satellite

m = Mass of the earth

r = The distance of the satellite from the earth

R = Radius of earth

Calculation:

From given formula,

  $\therefore PE\propto -\frac{1}{r}$

From the given formula, distance of the satellite increases as the satellite near to the earth. Since the potential is negative. So, the potential energy is decreases from position A to position D

Conclusion:

Therefore, potential energy of satellite, from greatest to least is, D > C > B > A

To determine

To rank: Total energy ( E ) of satellite, from greatest to least.

Answer to Problem 22A

Total energy ( E ) of satellite, from greatest to least is, A = B = C = D

Explanation of Solution

Given:

The position of satellite in elliptical orbit are shown below.

  

Formula:

The total energy ( E ) of the satellite in elliptical orbit is,

  $E=PE+KE$

Where,

PE = Potential energy of the satellite

KE = Kinetic energy of the satellite

Calculation:

According to conservation of energy, the total energy is constant in any process. When satellite move around the earth in elliptical orbit, depending on its position, velocity of the satellite get changed. When particle is move near to the earth due to gravitational pull, speed of the satellite increases. That increase in the speed of the satellite responsible for the higher kinetic energy. Similarly, as particle move away from the earth, its potential energy increases. Since the total energy of the satellite during its journey is remain constant.

Conclusion:

Therefore, total energy ( E ) of satellite, from greatest to least is, A = B = C = D

To determine

To rank: Acceleration of satellite, from greatest to least.

Answer to Problem 22A

Acceleration of satellite, from greatest to least is, A > B > C > D

Explanation of Solution

Given:

The position of satellite in elliptical orbit are shown below.

  

Formula:

The acceleration of the satellite in elliptical orbit is,

F= ma

Where,

F = Force acting on the satellite

m = mass of the satellite

a =Acceleration of the satellite

Calculation:

It is known from the part (a), as the distance between the satellite and Earth decreases gravitational force acting on the satellite increases. Since the acceleration of the satellite is directly proportional to the gravitational force on the satellite. Therefore, as distance between earth and satellite decreases, acceleration of the satellite increases.

Conclusion:

Therefore, acceleration of satellite, from greatest to least is, A > B > C > D