Chapter 8.15 - When Initial velocity is Greater than or Less than Escape velocity

In the previous section, we saw escape velocity
• In this section we will see what happens when the launch velocity is less than or greater than escape velocity

When the launch velocity is greater than the escape velocity
1. Suppose that, an object is launched with a velocity of v_i
• Let v_i be greater than v_e
2. Total energy on the surface of the earth = $\mathbf\small{\frac{1}{2}mv_i^2-\frac{G\,M_E\,m}{R_E}}$
3. Total energy at the final position = $\mathbf\small{\frac{1}{2}mv_f^2+0}$
• Note that, v_f will not be zero when v_i is greater than v_e. We saw the reason in the previous section
4. Equating the two energies, we get: $\mathbf\small{\frac{1}{2}mv_i^2-\frac{G\,M_E\,m}{R_E}=\frac{1}{2}mv_f^2}$
$\mathbf\small{\Rightarrow \frac{1}{2}v_i^2-\frac{G\,M_E}{R_E}=\frac{1}{2}v_f^2}$
5. But we have: $\mathbf\small{v_e=\sqrt{\frac{2GM_E}{R_E}}}$
$\mathbf\small{\Rightarrow v_e^2=\frac{2GM_E}{R_E}}$
$\mathbf\small{\Rightarrow \frac{v_e^2}{2}=\frac{GM_E}{R_E}}$
6. So we can replace the second term in (4). We get:
$\mathbf\small{\frac{1}{2}v_i^2-\frac{v_e^2}{2}=\frac{1}{2}v_f^2}$
$\mathbf\small{\Rightarrow v_i^2-v_e^2=v_f^2}$
Thus we get:
Eq.8.20: $\mathbf\small{v_f=\sqrt{v_i^2-v_e^2}}$
7. Thus we can easily calculate vf
■ Once the object is out of the gravitational field, there will not be any force acting on it. So that object will begin to move with a constant velocity of vf

When the launch velocity is less than the escape velocity
1. Suppose that, an object is launched with a velocity of v_i
• Let v_i be less than v_e
2. Total energy on the surface of the earth = $\mathbf\small{\frac{1}{2}mv_i^2-\frac{G\,M_E\,m}{R_E}}$
3. Total energy at the final position:
(i) In this case, the object does not escape from the gravitational field. So it's potential energy does not become zero at the final position
(ii) Let it rise to a maximum height h above the surface of the earth
• Then the potential energy at the final position is $\mathbf\small{-\frac{G\,M_E\,m}{(R_E+h)}}$
(iii) The object continues to rise until it's velocity becomes zero
• h is the height attained at the instant when velocity becomes zero
• So the final kinetic energy is zero
(iv) Thus we can write:
Total energy at the final position = $\mathbf\small{-\frac{G\,M_E\,m}{(R_E+h)}+0}$
4. Equating the energies in (2) and (3), we get:
$\mathbf\small{\frac{1}{2}mv_i^2-\frac{G\,M_E\,m}{R_E}=-\frac{G\,M_E\,m}{(R_E+h)}}$
$\mathbf\small{\Rightarrow \frac{1}{2}v_i^2=\frac{G\,M_E}{R_E}-\frac{G\,M_E}{(R_E+h)}}$
$\mathbf\small{\Rightarrow \frac{1}{2}v_i^2=G\,M_E\left(\frac{1}{R_E}-\frac{1}{(R_E+h)}\right)}$
$\mathbf\small{\Rightarrow \frac{1}{2}v_i^2=G\,M_E\left(\frac{h}{R_E(R_E+h)}\right)}$
$\mathbf\small{\Rightarrow \frac{1}{2}v_i^2=\frac{G\,M_E}{R_E^2}\left(\frac{h}{(1+\frac{h}{R_E})}\right)}$
$\mathbf\small{\Rightarrow \frac{1}{2}v_i^2=\frac{G\,M_E}{R_E^2}\left(\frac{h}{(1+\frac{h}{R_E})}\right)}$
$\mathbf\small{\Rightarrow v_i^2=2g\left(\frac{h}{(1+\frac{h}{R_E})}\right)}$
$\mathbf\small{\Rightarrow v_i^2=\frac{2\,g\,h}{(1+\frac{h}{R_E})}}$
$\mathbf\small{\Rightarrow v_i^2+v_i^2\frac{h}{R_E}=2\,g\,h}$
$\mathbf\small{\Rightarrow v_i^2\,R_E+v_i^2\,h=2\,g\,h\,R_E}$
$\mathbf\small{\Rightarrow v_i^2\,R_E=(2\,g\,R_E-v_i^2)h}$
$\mathbf\small{\Rightarrow h=\frac{v_i^2\,R_E}{2\,g\,R_E-v_i^2}}$
Thus we get Eq.8.21: $\mathbf\small{h=\frac{v_i^2}{2\,g-\frac{v_i^2}{R_E}}}$
• This is the maximum height that can be achieved when the launch velocity is less than the escape velocity 

Now we will see some solved examples
Solved example 8.38
A body is projected upwards with a velocity of (4 × 11.2) km s^-1 from the surface of the earth. What will be the velocity of the body when it escapes from the gravitational field of the earth?
Solution:
• In this problem the launch velocity is greater than escape velocity. So we will use Eq.8.20:
$\mathbf\small{v_f=\sqrt{v_i^2-v_e^2}}$
• Substituting the values, we get: $\mathbf\small{v_f=\sqrt{(4 \times 11.2)^2-11.2^2}=\sqrt{15 \times 11.2^2}=11.2\sqrt{15}}$

Solved example 8.39
A body is projected upwards from the surface of the earth with a velocity equal to one fourth the escape velocity. What is the maximum height that the body will achieve?
Solution:
1. In this problem, the launch velocity is less than escape velocity
• In such cases, we can use Eq.8.21: $\mathbf\small{h=\frac{v_i^2}{2\,g-\frac{v_i^2}{R_E}}}$
2. But it is more convenient to start from the basics. That is., we start by equating the energies:
$\mathbf\small{\frac{1}{2}mv_i^2-\frac{G\,M_E\,m}{R_E}=-\frac{G\,M_E\,m}{(R_E+h)}}$
$\mathbf\small{\Rightarrow \frac{1}{2}v_i^2=\frac{G\,M_E}{R_E}-\frac{G\,M_E}{(R_E+h)}}$
$\mathbf\small{\Rightarrow \frac{1}{2}v_i^2=\frac{G\,M_E}{R_E}-\frac{G\,M_E}{R_E(1+\frac{h}{R_E})}}$
3. But we have: $\mathbf\small{v_e=\sqrt{\frac{2GM_E}{R_E}}}$
$\mathbf\small{\Rightarrow v_e^2=\frac{2GM_E}{R_E}}$
$\mathbf\small{\Rightarrow \frac{v_e^2}{2}=\frac{GM_E}{R_E}}$
4. So the equation in (2) becomes:
$\mathbf\small{\frac{1}{2}v_i^2=\frac{v_e^2}{2}-\frac{v_e^2}{2(1+\frac{h}{R_E})}}$
5. Given that launch velocity is equal to one fourth the escape velocity. So we get:
$\mathbf\small{\frac{1}{32}v_e^2=\frac{v_e^2}{2}-\frac{v_e^2}{2(1+\frac{h}{R_E})}}$
$\mathbf\small{\Rightarrow \frac{1}{32}=\frac{1}{2}-\frac{1}{2(1+\frac{h}{R_E})}}$
$\mathbf\small{\Rightarrow \frac{1}{16}=1-\frac{1}{(1+\frac{h}{R_E})}}$
$\mathbf\small{\Rightarrow \frac{1}{(1+\frac{h}{R_E})}=\frac{15}{16}}$
$\mathbf\small{\Rightarrow (1+\frac{h}{R_E})=\frac{16}{15}}$
$\mathbf\small{\Rightarrow \frac{h}{R_E}=\frac{1}{15}}$
$\mathbf\small{\Rightarrow h=\frac{R_E}{15}}$

Solved example 8.40
A body has to reach a height R_E above the surface of the earth. What is the required launch velocity?
Solution:
1. In this problem, a definite target height is given. So the launch velocity is less than escape velocity. Other wise the object will escape from the gravitational field of the earth
• In such cases, we can use Eq.8.21: $\mathbf\small{h=\frac{v_i^2}{2\,g-\frac{v_i^2}{R_E}}}$
2. But it is more convenient to start from the basics. That is., we start by equating the energies:
$\mathbf\small{\frac{1}{2}mv_i^2-\frac{G\,M_E\,m}{R_E}=-\frac{G\,M_E\,m}{(R_E+h)}}$
$\mathbf\small{\Rightarrow \frac{1}{2}v_i^2=\frac{G\,M_E}{R_E}-\frac{G\,M_E}{(R_E+h)}}$
$\mathbf\small{\Rightarrow \frac{1}{2}v_i^2=\frac{G\,M_E}{R_E}-\frac{G\,M_E}{R_E(1+\frac{h}{R_E})}}$
3. But we have: $\mathbf\small{v_e=\sqrt{\frac{2GM_E}{R_E}}}$
$\mathbf\small{\Rightarrow v_e^2=\frac{2GM_E}{R_E}}$
$\mathbf\small{\Rightarrow \frac{v_e^2}{2}=\frac{GM_E}{R_E}}$
4. So the equation in (2) becomes:
$\mathbf\small{\Rightarrow \frac{1}{2}v_i^2=\frac{v_e^2}{2}-\frac{v_e^2}{2(1+\frac{h}{R_E})}}$
5. Given that h = R_E. So we get:
$\mathbf\small{\frac{1}{2}v_i^2=\frac{v_e^2}{2}-\frac{v_e^2}{2(1+\frac{R_E}{R_E})}=\frac{v_e^2}{2}-\frac{v_e^2}{4}}$
$\mathbf\small{\Rightarrow \frac{1}{2}v_i^2=\frac{v_e^2}{2}-\frac{v_e^2}{4}=\frac{v_e^2}{4}}$
$\mathbf\small{\Rightarrow v_i=\frac{v_e}{\sqrt{2}}}$
6. But we have: $\mathbf\small{v_e=\sqrt{\frac{2GM_E}{R_E}}}$
So the result in (4) becomes: $\mathbf\small{v_i=\sqrt{\frac{2GM_E}{R_E}} \times\frac{1}{\sqrt{2}}}$
$\mathbf\small{\Rightarrow v_i=\sqrt{\frac{GM_E}{R_E}}}$

Solved example 8.41
A rocket is fired vertically with a speed of 5 km s^-1 from the earth’s surface. How far from the earth does the rocket go before returning to the earth ? Mass of the earth = 6.0 × 10²⁴ kg; mean radius of the earth = 6.4 × 10⁶  m; G = 6.67 × 10^-11 N m² kg^-2
Solution:
1. In this problem, the launch velocity is less than escape velocity. The actual values are also given
• In such cases, we can use Eq.8.21: $\mathbf\small{h=\frac{v_i^2}{2\,g-\frac{v_i^2}{R_E}}}$
2. Substituting the values, we get:
$\mathbf\small{h=\frac{5000^2}{2(9.81)-\frac{5000^2}{6400000}}=1590963.33}$ m
3. So distance from the center of the earth = (R_E+h) = (6400000 + 1590963.33) 
= 7990963.33 m = 8 × 10⁶ m

Solved example 8.42
The escape speed of a projectile on the earth’s surface is 11.2 km s^-1. A body is projected out with thrice this speed. What is the speed of the body far away from the earth? Ignore the presence of the sun and other planets.
Solution:
• In this problem the launch velocity is greater than escape velocity. So we will use Eq.8.20:
$\mathbf\small{v_f=\sqrt{v_i^2-v_e^2}}$

• Substituting the values, we get: $\mathbf\small{v_f=\sqrt{(3 \times 11.2)^2-11.2^2}=\sqrt{8 \times 11.2^2}=11.2\sqrt{8}=31.7}$ km s^-1

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• In the next section we will see earth of satellites

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Chapter 8.14 - Escape Velocity

In the previous section, we saw gravitational potential
• In this section we will see Escape velocity

1. In fig.8.47 below, an object of mass m is being launched from the surface of the earth

Fig.8.47

• Let the initial velocity be v_i
• Then the initial kinetic energy K_i = $\mathbf\small{\frac{1}{2}mv_i^2}$
2. We already know the initial potential energy U_i
• That is., the potential energy when the object is at the surface of the earth
• It is given by: U_i = $\mathbf\small{-\frac{G\,M_E\,m}{R_E}}$
3. So the initial total energy = K_i + U_i = $\mathbf\small{\frac{1}{2}mv_i^2-\frac{G\,M_E\,m}{R_E}}$
4. Now we want the final values
• We want the object to escape from the gravitational field of the earth
• We know that, as the object rises, the speed goes on decreasing
    ♦ If v_i is large the object will rise to a larger height
    ♦ If v_i is small, the object will rise only to a smaller height
    ♦ If v_i is large enough, the object will rise to such a height that, it is out of the gravitational field
• The height at which the object is out of the gravitational field is our final position. This is shown in fig.8.48 below:

Fig.8.48

5. If v_i is more than 'what is sufficient', the object will continue to rise even after escaping from the field
• We do not want this to happen. What we want is to ‘just escape’. We do not want the object to travel further
6. That means, at the instant when the object is out of the field, the velocity must be zero
• This implies that, at the instant when the object is out of the field, the kinetic energy must be zero
7. We can write:
At the final position, v_f = 0 ⇒ K_f = 0

8. There is also another implication arising from 'just escape'. We can write it in 6 steps:
(i) If the initial velocity v_i is more than ‘what is sufficient’, the object will travel further even after getting out of the field
(ii) We do not want this to happen. We want to ‘just escape’
(iii) So v_i must be just equal to ‘what is sufficient’ to ‘just escape’
(iv) We call that velocity: The escape velocity
• It is denoted as v_e
(v) The initial velocity v_i must be equal to v_e. So we can write:
Initial kinetic energy K_i = $\mathbf\small{\frac{1}{2}mv_e^2}$ 
(vi) Thus from (3), we get:
Total initial energy = K_i + U_i = $\mathbf\small{\frac{1}{2}mv_e^2-\frac{G\,M_E\,m}{R_E}}$
9. Next we calculate the potential energy at the final instant
• The ‘instant when the object is out of the field’ implies that, at that instant, there is no influence of the field
• We will have a potential energy only if there is an influence of the field
• So at the final position, there is no potential energy
That is., U_f = 0
10. Thus using the results in (7) and (9), we can write:
Total energy at the final point = (K_f + U_f) = (0 + 0) = 0 
11. The total initial energy must be equal to the total final energy
• Using the results in (8) and (10), we get: $\mathbf\small{\frac{1}{2}mv_e^2-\frac{G\,M_E\,m}{R_E}=0}$
$\mathbf\small{\Rightarrow \frac{1}{2}v_e^2=\frac{G\,M_E}{R_E}}$
• Thus we get: Eq.8.17: $\mathbf\small{v_e=\sqrt{\frac{2G\,M_E}{R_E}}}$
12. Multiplying both numerator and denominator by √R_E, we get: $\mathbf\small{v_e=\sqrt{\frac{2G\,M_E\,R_E}{R_E^2}}}$
• Thus we get: Eq.8.18: $\mathbf\small{v_e=\sqrt{2gR_E}}$
(∵ $\mathbf\small{\frac{2G\,M_E}{R_E^2}=g}$)
13. Inputting the values in Eq.8.18, we get:
$\mathbf\small{v_e=\sqrt{2\times 9.81 \times 6.371 \times 10^6}=11.18 \times 10^3}$ ms^-1
14. An important note:
• Throw some thing upwards with initial velocity v_e
    ♦ It will escape from the gravitational field of the earth
    ♦ It will escape without any need for 'propulsion by burning fuel'
• Throw some thing upwards with initial velocity less than v_e
    ♦ It will escape only if 'propulsion by burning fuel' is supplied   
15. Eq.8.17 shows that, the escape velocity depends only on the mass and radius of the earth
We can write the escape velocity for any planet P:
Eq.8.19: $\mathbf\small{v_{e(P)}=\sqrt{\frac{2G\,M_P}{R_P}}}$
Where:
    ♦ $\mathbf\small{v_{e(P)}}$ is the escape velocity of the planet P
    ♦ $\mathbf\small{M_P}$ is the mass of the planet P
    ♦ $\mathbf\small{R_P}$ is the radius of the planet P

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Now we will see some solved examples

Solved example 8.33
Calculate the escape velocity from the surface of a planet of mass 14.8 × 10²² kg. Radius of the planet is 3.48 × 10⁶ m
Solution:
• We have Eq.8.19: $\mathbf\small{v_{e(P)}=\sqrt{\frac{2G\,M_P}{R_P}}}$
• Substituting the values, we get:
$\mathbf\small{v_{e(P)}=\sqrt{\frac{2\times 6.67 \times 10^{-11}\times 14.8 \times 10^{22}}{3.48 \times 10^{6}}}}$ = 2381.9 m s^-1 = 2.38 km s^-1

Solved example 8.34
Mass of Jupiter is 318 times the mass of earth. Radius of Jupiter is 11.2 times the radius of earth. If the escape velocity of earth is 11.2 km s^-1, calculate the escape velocity of Jupiter
Solution:
1. For earth, we have: $\mathbf\small{v_{e(E)}=\sqrt{\frac{2G\,M_E}{R_E}}}$  
• For Jupiter, we have: $\mathbf\small{v_{e(J)}=\sqrt{\frac{2G\,M_J}{R_J}}}$
2. Taking ratio of squares, we get:
$\mathbf\small{\frac{v_{e(E)}^2}{v_{e(J)}^2}=\frac{2G\,M_E}{R_E}\times \frac{R_J}{2G\,M_J}=\frac{M_E}{M_J}\times \frac{R_J}{R_E}}$
$\mathbf\small{\Rightarrow \frac{v_{e(E)}^2}{v_{e(J)}^2}=\frac{M_E}{318M_E}\times \frac{11.2R_E}{R_E}=\frac{11.2}{318}}$
$\mathbf\small{\Rightarrow v_{e(J)}^2=\frac{318}{11.2}v_{e(E)}^2}$
$\mathbf\small{\Rightarrow v_{e(J)}^2=\frac{318}{11.2}\times 11.2^2=318 \times 11.2=3561.6}$
$\mathbf\small{\Rightarrow v_{e(J)}=\sqrt{3561.6}=59.7}$ km s^-1 

Solved example 8.35
The earth is assumed to be a sphere of radius R_E. A platform is arranged at a height of R_E from the surface of the earth. The escape velocity of a body from this platform is (k×v_e). Where v_e is the escape velocity from the surface of the earth. Find the value of k
Solution:
1. The platform is at a distance of 2RE from the center of the earth
• So potential energy of the body when it is on the platform = $\mathbf\small{-\frac{GM_Em}{2R_E}}$
• Where m is the mass of the body
2. The body is launched with an initial velocity of kve from the platform
• So kinetic energy of the body when it is at the platform = $\mathbf\small{\frac{1}{2}m(kv_e)^2=\frac{1}{2}mk^2v_e^2}$
3. So total energy at the platform = $\mathbf\small{-\frac{GM_Em}{2R_E}+\frac{1}{2}mk^2v_e^2}$
4. Total energy when the body has just escaped from earth's gravitational field = 0
5. Applying law of conservation of energy, the results in (3) and (4) must be equal
• So we can write: $\mathbf\small{-\frac{GM_Em}{2R_E}+\frac{1}{2}mk^2v_e^2=0}$
$\mathbf\small{\Rightarrow -\frac{GM_E}{R_E}+k^2v_e^2=0}$
$\mathbf\small{\Rightarrow \frac{GM_E}{R_E}=k^2v_e^2}$
6. Multiplying the numerator and denominator of the left side by 2, we get:
$\mathbf\small{\frac{2GM_E}{2R_E}=k^2v_e^2}$
$\mathbf\small{\Rightarrow \frac{v_e^2}{2}=k^2v_e^2}$
(∵ $\mathbf\small{\frac{2GM_E}{R_E}=v_e^2}$)
$\mathbf\small{\Rightarrow \frac{1}{2}=k^2}$
$\mathbf\small{\Rightarrow k=\frac{1}{\sqrt{2}}}$

Solved example 8.36
We know that, to escape from the gravitational field of the earth, a body must be given a 'minimum initial kinetic energy'. Which of the following is equal to that kinetic energy?
$\mathbf\small{\frac{mgR_E}{2},\;2mgR_E\;,\;mgR_E\;,\;\frac{3mgR_E}{2}}$
Solution:
1. The 'minimum initial kinetic energy' will be equal to $\mathbf\small{\frac{1}{2}mv_e^2}$
2. But v_e is equal to $\mathbf\small{\sqrt{2gR_E}}$
3. Substituting in (1), we get:
'Minimum initial kinetic energy' = $\mathbf\small{\frac{1}{2}m(\sqrt{2gR_E})^2}$
= $\mathbf\small{\frac{1}{2}m \times(2gR_E)=mgR_E}$

Solved example 8.37
Two uniform solid spheres of equal radii R, but mass M and 4 M have a center to center separation 6R, as shown in fig. 8.49(a). The two spheres are held fixed. A projectile of mass m is projected from the surface of the sphere of mass M directly towards the center of the second sphere. Obtain an expression for the minimum initial speed v_i of the projectile so that it reaches the surface of the second sphere

Fig.8.49

Solution:
1. The mass M will exert an attractive force on m
• The magnitude of this attractive force will be $\mathbf\small{\frac{GMm}{r^2}}$
    ♦ Where r is the distance of m from the center of M
    ♦ This is shown in fig.b
• This force acts towards the left
2. The mass 4M will also exert an attractive force on m
• The magnitude of this attractive force will be $\mathbf\small{\frac{4GMm}{(6R-r)^2}}$
    ♦ Where r is the distance of m from the center of 4M
• This force acts towards the right
3. The two forces act in opposite directions
The magnitudes of the forces depend on the value of r
4. Let the magnitudes become equal when the m is at N
• Then we have: $\mathbf\small{\frac{GMm}{r^2}=\frac{4GMm}{(6R-r)^2}}$
$\mathbf\small{\Rightarrow \frac{1}{r^2}=\frac{4}{(6R-r)^2}}$
$\mathbf\small{\Rightarrow \frac{1}{r}=\frac{\pm 2}{(6R-r)}}$
• Solving this equation, we get: r = 2R or -6R
• For our present problem, -6R is not acceptable. So we take r = 2R
• We can write: At N, the mass m will experience a zero net force
5. Our aim is to make the mass m to reach N
• We must be able to accomplish this by supplying the 'least possible' external energy
6. The external energy is in the form of kinetic energy
• So we can write:
We must be able to accomplish the task in (5) using the 'least possible' launch velocity
• The launch velocity is the initial velocity vi
• So we must find the least possible v_i so that, m will 'just reach' N
• 'Just reach' indicates that, the velocity of m will be zero when it reaches N 
7. So the velocity of m will become zero when it reaches N
• Even though the velocity becomes zero, when it reaches N, the larger mass 4M will pull the mass and so, the m can reach 4M
8. Thus we can write:
The kinetic energy at N is zero
9. We will consider the journey from M to N
• K_i and U_i are the energies of m when it is at the surface of M
• K_f and U_f are the energies of m when it is at N
10. Next, we will write the values of those energies:
• We have: K_i = $\mathbf\small{\frac{1}{2}mv_i^2}$ 
• U_i will be the sum of two items:
    ♦ Contribution from M
    ♦ Contribution from 4M
• So we get: U_i = $\mathbf\small{-\frac{GMm}{R}+-\frac{4GMm}{5R}}$

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• Why do we say that the contribution from 4M is $\mathbf\small{-\frac{4GMm}{5R}}$?

The answer can be written in 4 steps:
(i) The 4M mass creates a 'gravitational potential' around itself in all directions
(We have seen the details in the previous section)
(ii) At a distance of 5R, this potential is equal to $\mathbf\small{-\frac{4GM}{5R}}$
(iii) This is the potential experienced by unit mass
(iv) So a mass m will experience a potential of $\mathbf\small{-\frac{4GMm}{5R}}$

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11. So total initial energy = K_i + U_i = $\mathbf\small{\frac{1}{2}mv_i^2-\frac{GMm}{R}-\frac{4GMm}{5R}}$
12. Next we take up the final energies
We know that U_f = 0
13. U_f will be the sum of two items:
    ♦ Contribution from M
    ♦ Contribution from 4M
• So we get: U_f = $\mathbf\small{-\frac{GMm}{2R}+-\frac{4GMm}{4R}}$
14. So total final energy = K_f + U_f = $\mathbf\small{0-\frac{GMm}{2R}-\frac{4GMm}{4R}}$
15. The results in (11) and (14) must be equal. So we can write:
$\mathbf\small{\frac{1}{2}mv_i^2-\frac{GMm}{R}-\frac{4GMm}{5R}=\frac{GMm}{2R}-\frac{4GMm}{4R}}$
• Dividing both sides by m, we get:
$\mathbf\small{\frac{1}{2}v_i^2-\frac{GM}{R}-\frac{4GM}{5R}=\frac{GM}{2R}-\frac{4GM}{4R}}$
• The denominators of the ^GM⁄_R terms are 1, 2, 4 and 5. The LCM is 20
• Multiplying both numerators and denominators of those terms by 20, we get:
$\mathbf\small{\frac{1}{2}v_i^2-\frac{20GM}{20R}-\frac{16GM}{20R}=-\frac{10GM}{20R}-\frac{20GM}{20R}}$
$\mathbf\small{\Rightarrow \frac{1}{2}v_i^2=\frac{6GM}{20R}}$
$\mathbf\small{\Rightarrow v_i^2=\frac{3GM}{5R}}$
$\mathbf\small{\Rightarrow v_i=\sqrt{\frac{3GM}{5R}}}$

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Now we know the basics about escape velocity. We must always keep the following 5 points in mind:

1. Escape velocity does not depend on the angle of projection. That is., the direction of the velocity vector is not important. 
• This is because, we are considering only the 'kinetic energy due to the velocity'
• But most rockets are launched towards the east direction
• The reason can be explained using fig.8.50 below:

Fig.8.50

(i) The blue sphere represents the earth. The yellow line is the axis
(ii) The direction of spin is indicated by the curved arrow. This curved arrow points towards the east
• That is., the earth spins towards the east
(iv) So a rocket is launched towards the east. This will give an additional velocity to the rocket
2. From Eq.8.17, we see that, the escape velocity depends upon mass and radius. G is an universal constant
• So value of escape velocity is different for different planets
3. Different molecules in the gaseous atmosphere will be having different velocities. We take an average velocity for calculations. This average velocity is called 'root mean square velocity'.
(i) Consider any planet
(ii) Note down the root mean square velocity of the molecules in the atmosphere of that planet
(iii) If this velocity is less than the escape velocity of that planet, the molecules cannot escape
(iv) In that case, an atmosphere will be present for that planet
• For moon, the escape velocity is very low: 2.3 km s^-1
    ♦ So the molecules are able to escape into space, leaving the moon with no atmosphere
• For Jupiter, the escape velocity is very high: 60 km s^-1
    ♦ So the molecules are not able to escape. Jupiter has a thick atmosphere
4. We saw that, escape velocity is the initial velocity at the time of launch from the surface of planet
(i) As the object rises, it's velocity decreases
(ii) Finally, when it reaches outer space where there is no influence of gravity, the velocity becomes zero
• The revers steps are also true:
• Drop an object from outer space where there is no influence of gravity
('Drop' indicates that initial velocity is zero)
(i) As the objects descends, the velocity increases
(ii) Finally, when it hits the surface, the velocity will be equal to the escape velocity
5. In calculating the escape velocity, 'air resistance' and 'gravitational effect due to other celestial bodies' are neglected

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• In the next section we will see what happens if the speed of launch is greater than or less than the escape velocity

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Chapter 8.13 - Gravitational Potential

In the previous section, we saw gravitational potential energy
• We derived Eq.8.15: $\mathbf\small{U_r=-\frac{G\,M_E\,m}{r}}$
• In this section we will see how it is related to our old equation:
Gravitational Potential energy = mgh
• Later in this section we will also see 'gravitational potential'

1. Consider a point mass m
2. Let it be placed on the surface of the earth
• Then it's distance from the center O of the earth is R_E
• So it's potential energy is given by: $\mathbf\small{U_{R_E}=-\frac{G\,M_E\,m}{R_E}}$
3. Let it be taken to a height h above the surface of the earth
• Then it's distance from the center O of the earth is R_E+h
• So now it's potential energy is given by: $\mathbf\small{U_{(R_E+h)}=-\frac{G\,M_E\,m}{(R_E+h)}}$
4. Difference in potential energies = Final energy - Initial energy
= $\mathbf\small{U_{(R_E+h)}-U_{R_E}}$
= $\mathbf\small{-\frac{G\,M_E\,m}{(R_E+h)}-\,-\frac{G\,M_E\,m}{R_E}}$
= $\mathbf\small{-\frac{G\,M_E\,m}{(R_E+h)}+\frac{G\,M_E\,m}{R_E}}$
= $\mathbf\small{G\,M_E\,m \left(\frac{1}{R_E}-\frac{1}{(R_E+h)}\right)}$
= $\mathbf\small{G\,M_E\,m \left(\frac{R_E+h-R_E}{R_E(R_E+h)}\right)}$
= $\mathbf\small{G\,M_E\,m \left(\frac{h}{R_E(R_E+h)}\right)}$
= $\mathbf\small{G\,M_E\,m \left(\frac{h}{R_E^2(1+\frac{h}{R_E})}\right)}$
= $\mathbf\small{\frac{G\,M_E}{R_E^2} \left(\frac{mh}{(1+\frac{h}{R_E})}\right)}$
= $\mathbf\small{\frac{G\,M_E}{R_E^2} \left(\frac{mh}{(1)}\right)}$ (∵ h is small, $\mathbf\small{\frac{h}{R_E}}$ can be ignored)
= $\mathbf\small{m|\vec{g}|h}$ (∵ $\mathbf\small{\frac{G\,M_E}{R_E^2}=|\vec{g}|}$)
5. Thus we get the old relation:
Work done to raise an object of mass m from the surface of earth to a height h = $\mathbf\small{m|\vec{g}|h}$

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• So we have completed a discussion on gravitational potential energy

• Next we have to learn about gravitational potential. We can write about it in steps:
1. In fig.8.43 below, two bodies A and B are placed at P₁ and P₂

Fig.8.43

• Their masses are m_A and m_B respectively
• The center to center distance is r
2. So we have a system consisting of two masses m_A and m_B
• The gravitational potential energy of this system is given by: $\mathbf\small{U=-\frac{G\,m_A\,m_B}{r}}$
3. In normal cases, the masses m_A and m_B do not change
• But the positions can change:
    ♦ A can move away from P₁
    ♦ B can move away from P₂
    ♦ Both A and B can move away from their respective positions P₁ and P₂
• If any of those ‘changes in positions’ happen, the energy of the system will change
• So ‘positions of objects’ is important for finding the potential energy  
4. In the above fig., if m_B = 1 kg, the energy of the system will be equal to $\mathbf\small{U=-\frac{G\,m_A}{r}}$
• We can write: The body A is able to produce a gravitational potential of $\mathbf\small{-\frac{G\,m_A}{r}}$ joules at a distance of r
5. Similarly, if m_A = 1 kg, the energy of the system will be equal to $\mathbf\small{U=-\frac{G\,m_B}{r}}$
• We can write: The body B is able to produce a gravitational potential of $\mathbf\small{-\frac{G\,m_B}{r}}$ joules at a distance of r
6. We can define gravitational field in 5 steps:
(i) Consider a body A. Let it's mass be m_A
• There will be a gravitational field around A
(ii) A mass of 1 kg is initially at infinity
(iii) We want to bring this 1 kg mass into the field of A
• We want to place this 1 kg mass at a distance of r from A
(iv) For that, we have to do a work of $\mathbf\small{-\frac{G\,m_A}{r}}$ joules
• This much work will be stored in that 1 kg mass
• This much work is called gravitational potential (at distance r) created by A
(v) Gravitational potential is denoted by the letter V. Since it is an energy, it is a scalar quantity 
• So we can write:
Eq.8.16: $\mathbf\small{V_A=-\frac{G\,m_A}{r}}$
    ♦ The subscript ‘A’ indicates that, it the gravitational potential created by A 

The following solved example will help us to fully understand this concept
Solved example 8.29
Four equal masses m are placed at the four corners of a square ABCD. The side of the square is a. What is the gravitational potential energy of the system? Also find the gravitational potential at the center of the system
Solution:
Part (i):
1. Fig.8.44(a) below shows the arrangement

Fig.8.44

2. We have to consider one pair at a time
• Consider the pair A-B
• The potential energy due to this pair is given by: $\mathbf\small{U_{A,B}=-\frac{Gm^2}{a}}$
3. Along the periphery, there are 3 more pairs like this. All the four pairs along the periphery are identical. So we can write:
Total gravitational potential energy of the pairs along the periphery
= $\mathbf\small{U_{AB}\;+U_{B,C}\;+U_{C,D}\;+U_{D,A}}$
= $\mathbf\small{-\frac{Gm^2}{a}\;+-\frac{Gm^2}{a}\;+-\frac{Gm^2}{a}\;+-\frac{Gm^2}{a}}$
= $\mathbf\small{-\frac{4Gm^2}{a}}$
4. Next we consider the pair A-C along the diagonal
• The distance between the two masses in this pair = √2 a
    ♦ This is shown in fig.b
• So the potential energy due to this pair is given by: $\mathbf\small{U_{A,C}=-\frac{Gm^2}{\sqrt{2}\,a}}$
5. There is one more diagonal pair B-D like this. Both the diagonal pairs are identical. So we can write:
Total gravitational potential energy of the diagonal pairs
= $\mathbf\small{U_{A,C}\;+U_{B,D}}$
= $\mathbf\small{-\frac{Gm^2}{\sqrt{2}\,a}\;-\frac{Gm^2}{\sqrt{2}\,a}}$
= $\mathbf\small{-\frac{2Gm^2}{\sqrt{2}\,a}\;=-\frac{\sqrt{2}Gm^2}{\,a}}$
6. Thus we get:
Total gravitational potential energy of the system = $\mathbf\small{-\frac{4Gm^2}{a}\;+-\frac{\sqrt{2}Gm^2}{\,a}}$
= $\mathbf\small{-\frac{Gm^2}{a}(4+\sqrt{2})=-5.41\frac{Gm^2}{a}}$
Part (ii):
1. In this part we calculate the gravitational potential at the center
• We have to consider the potential created by each mass
• First we consider the mass at A
• It is at a distance of $\mathbf\small{\frac{a}{\sqrt{2}}}$ from the center O
    ♦ This is shown in fig.c
• So we get: $\mathbf\small{V_A=-\frac{Gm}{\frac{a}{\sqrt{2}}}=-\frac{\sqrt{2}Gm}{a}}$
2. There are three more masses. All of them are at the same distance of $\mathbf\small{\frac{a}{\sqrt{2}}}$ from the center. So all four masses will create the same potential
• Thus we get:
Total potential at the center of the square
= $\mathbf\small{V_A\;+V_B\;+V_C\;+V_D}$
= $\mathbf\small{-\frac{\sqrt{2}Gm}{a}\;+-\frac{\sqrt{2}Gm}{a}\;+-\frac{\sqrt{2}Gm}{a}\;+-\frac{\sqrt{2}Gm}{a}}$
= $\mathbf\small{-\frac{4\sqrt{2}Gm}{a}}$

Solved example 8.30
Three particles of masses m, 2m and 4m are placed at the corners of an equilateral triangle of side a
(i) Calculate the potential energy of the system
(ii) Work done on the system if all the sides are changed from a to 2a
Assume that the potential energy is zero when the sides are infinity
Solution:
Part (i):
1. Fig.8.45 below shows the arrangement

Fig.8.45

2. We have to consider one pair at a time
• Consider the pair A-B
    ♦ The potential energy due to this pair is given by: $\mathbf\small{U_{A,B}=-\frac{2Gm^2}{a}}$
• Consider the pair B-C
    ♦ The potential energy due to this pair is given by: $\mathbf\small{U_{B,C}=-\frac{8Gm^2}{a}}$
• Consider the pair C-A
    ♦ The potential energy due to this pair is given by: $\mathbf\small{U_{C,A}=-\frac{4Gm^2}{a}}$
3. Total gravitational potential energy of the system
= $\mathbf\small{U_{A,B}\;+U_{B,C}\;+U_{C,A}}$
= $\mathbf\small{-\frac{2Gm^2}{a}\;+-\frac{8Gm^2}{a}\;+-\frac{4Gm^2}{a}}$
= $\mathbf\small{-\frac{14Gm^2}{a}}$
Part (ii):
When the separation is 2a
1. We have to consider one pair at a time
• Consider the pair A-B
    ♦ The potential energy due to this pair is given by: $\mathbf\small{U_{A,B}=-\frac{2Gm^2}{2a}}$
• Consider the pair B-C
    ♦ The potential energy due to this pair is given by: $\mathbf\small{U_{B,C}=-\frac{8Gm^2}{2a}}$
• Consider the pair C-A
    ♦ The potential energy due to this pair is given by: $\mathbf\small{U_{C,A}=-\frac{4Gm^2}{2a}}$
2. Total gravitational potential energy of the system
= $\mathbf\small{U_{A,B}\;+U_{B,C}\;+U_{C,A}}$
= $\mathbf\small{-\frac{2Gm^2}{2a}\;+-\frac{8Gm^2}{a}\;+-\frac{4Gm^2}{2a}}$
= $\mathbf\small{-\frac{7Gm^2}{a}}$
3. To find the work done:
(i) Initially the particles are infinite distance apart. Then the energy of the system will be zero
(ii) Energy (U₁) of the system when the separation is a = $\mathbf\small{-\frac{14Gm^2}{a}}$
(iii) Energy (U₂) of the system when the separation is 2a = $\mathbf\small{-\frac{7Gm^2}{a}}$
(iv) Work done
= Change in energy
= Final energy - Initial energy
= U₂-U₁ = $\mathbf\small{-\frac{7Gm^2}{a}\;--\frac{14Gm^2}{a}}$
= $\mathbf\small{-\frac{7Gm^2}{a}\;+\frac{14Gm^2}{a}}$
= $\mathbf\small{\frac{7Gm^2}{a}}$

Solved example 8.31
An object is dropped from a height of 2R_E from the surface of the earth. Find the speed with which it will hit the surface of the earth. Neglect the effect of air resistance
Radius of the earth = R_E
Mass of the earth = M_E
Solution:
1. Let m be the mass of the object
• Potential energy of the object when it is at the surface of the earth = $\mathbf\small{-\frac{GM_Em}{R_E}}$
• Potential energy of the object when it is at a height of 2R_E = $\mathbf\small{-\frac{GM_Em}{R_E+2R_E}=-\frac{GM_Em}{3R_E}}$
2. So difference in potential energy = $\mathbf\small{-\frac{GM_Em}{3R_E}--\frac{GM_Em}{R_E}}$
= $\mathbf\small{\frac{GM_Em}{R_E}-\frac{GM_Em}{3R_E}=\frac{2GM_Em}{3R_E}}$
3. Since air resistance is neglected, we can write:
• The difference in potential energy will be converted into kinetic energy
4. Let v be the speed with which the object hits the ground
• Then it's kinetic energy at the instant of impact = $\mathbf\small{\frac{1}{2}mv^2}$
5. Equating the results in (2) and (4), we get: $\mathbf\small{\frac{1}{2}mv^2=\frac{2GM_Em}{3R_E}}$
$\mathbf\small{\Rightarrow v^2=\frac{4GM_E}{3R_E}}$
$\mathbf\small{\Rightarrow v=\sqrt{\frac{4GM_E}{3R_E}}=2\sqrt{\frac{GM_E}{3R_E}}}$

Solved example 8.32
In fig.8.46 below, two identical particles, each of mass m, are kept at rest at a distance d apart. They are allowed to move under the influence of their mutual gravitational force of attraction. What will be the speed of each when the distance between them is 0.5d

Fig.8.46

Solution:
1. Initially, the particles are at rest. So the only energy available initially is the potential energy which is equal to $\mathbf\small{-\frac{Gm^2}{d}}$
2. When the particles begin to move, there will be both potential energy and kinetic energy
3. Let v be the velocity of the particles at the instant when the distance between them is 0.5d
• Then the kinetic energy of the system at that instant = $\mathbf\small{\frac{1}{2}mv^2+\frac{1}{2}mv^2=mv^2}$ 
4. Potential energy of the system at that instant = $\mathbf\small{-\frac{Gm^2}{0.5d}=-\frac{2Gm^2}{d}}$  
5. So loss in potential energy
= Final potential energy - Initial potential energy
= $\mathbf\small{-\frac{2Gm^2}{d}--\frac{Gm^2}{d}}$
= $\mathbf\small{\frac{Gm^2}{d}-\frac{2Gm^2}{d}=-\frac{Gm^2}{d}}$
• The negative sign indicates that energy is lost 
6. 'Magnitude of this loss in potential energy' is equal to the 'kinetic energy of the system at that instant'
So equating the results in (3) and (5), we get: $\mathbf\small{\frac{Gm^2}{d}=mv^2}$
$\mathbf\small{\Rightarrow \frac{Gm}{d}=v^2}$
$\mathbf\small{\Rightarrow v=\sqrt{\frac{Gm}{d}}}$

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• In the next section we will see Escape velocity

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