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 vi
• Let vi be greater than ve
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, vf will not be zero when vi is greater than ve. 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 vi
• Let vi be less than ve
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 RE 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 = RE. 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 = (RE+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
• 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 vi
• Let vi be greater than ve
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, vf will not be zero when vi is greater than ve. 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 vi
• Let vi be less than ve
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 RE 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 = RE. 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 = (RE+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
• In the next section we will see earth of satellites
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