In the previous section, we completed a discussion on Geostationary satellites
• In this section we will see Polar satellites
1. In fig.7.59 below, the earth is shown as a blue sphere
• The axis of rotation of the earth is shown in yellow color
♦ It passes through the N and S poles
• The direction of rotation is indicated by the orange curved arrow
2. A satellite shown in red color orbits around the earth
• The orbit of the satellite is circular in shape. It is shown in green color
3. We see that, the orbit is vertical
• It passes through the N and S poles
• Such satellites are called Polar satellites
4. Polar satellites have low heights
• The value of h can be in between 500 and 800 km
5. We derived the relation in the previous section: $\mathbf\small{(R_E+h)=\left({\frac{T^2\,G\,M_E}{4\pi^2}}\right)^{1/3}}$
• h and T are the only variables. Both are in the numerators
• So, since the h is low, T will be also be low
6. Since T is low, we can say that:
• The polar satellites require only a lesser time to go around the earth once
• When the earth completes one rotation, the polar satellite would have completed many rotations
7. Consider a camera fixed on the polar satellite
• Since h is low, the camera can capture only a small 'area' of the earth
• This will be clear from fig.7.60 below:
(i) In fig.a, the satellite is at a smaller height from the surface of the earth
• In fig.b, the satellite is at a larger height
(ii) The angle 𝞱, which is the field of view, is same in both the cases
(iii) In fig.a, the camera is able to capture a width of AB
• In fig.b, the camera is able to capture a width of A'B'
(iv) We see that, A'B' is greater than AB
• It is obvious that, when height increases, more details can be captured
• But unfortunately, for polar satellites, the height is low
(v) Since the satellite is going in the vertical direction, the camera will be capturing a 'vertical strip'
• Because of the low h, this strip will be a narrow one
8. In fig.8.61 below, the surface of the earth is divided into strips
(i) In this position, the green orbit is above the blue strip
• So the camera is able to capture the details in the blue strip
(ii) Remember that, the green orbit is fixed. But the earth rotates about the N-S axis
• So various strips come below the orbit
(iii) When the satellite comes for the next round, the next green strip will be below the orbit
• So the camera gets the opportunity to capture the details in the green strip
(iv) This process continues and the camera is able to capture the details in all the strips
9. Remember that, the satellite is at a lower height
• So the pictures captured will have a good resolution
• We get good resolution pictures of the whole earth, strip by strip
• Because of this advantage, polar satellites are used in remote sensing, meteorology, and environmental studies of the earth
• In an earlier chapter 4, we have learnt about horizontal projectiles
• We saw how to calculate range, time of flight, path of the projectile etc., We will be using those details for our present discussion
1. Imagine person standing on top of a cliff. Let the height of his position above ground be 15 m
• He throws a stone horizontally with a velocity of 10 ms-1
• The stone will land at a horizontal distance (range) of 17.48 m from the foot of the cliff
• The path followed by the stone is the curve A in fig.8.62 below:
2. Let the person climb higher to 25 m
• He throws a stone horizontally with the same velocity of 10 ms-1
• This time the range is 22.58 m
• The path followed by the stone is the curve B in fig.8.62
3. Let the person be at the same height of 25 m
• He throws a stone horizontally with an increased velocity of 15 ms-1
• This time the range is 33.86 m
• The path followed by the stone is the curve C
4. Let the person climb much higher to 40 m
• He throws a stone horizontally with a high velocity of 35 ms-1
• This time the range is 99.95 m
• The path followed by the stone is the curve D
5. From the above 4 steps, we can write the following 2 points:
(i) Height and velocity are the only variables possible
• In other words, we can 'make adjustments' only on those two items. Other item 'g' is a constant
(ii) When height and velocity are increased, we can achieve a large range
• Note that, we have neglected the effect of air resistance in the calculations
6. Consider the point 5 (ii)
• We need to increase height and velocity
• This is exactly what we do in the case of a satellite
(i) The rocket carries the satellite to a very large height
(ii) At that height, small rockets are fired so that the satellite separates away from the main rocket
• The satellite separates away in a horizontal direction
(iii) So the motion of the satellite will be similar to the projectile motions that we saw in fig.8.62 above
7. This motion of the satellite has some advantages:
• Since height and velocity are large, the range will be very large
• At very large heights, the air will be so thin that, there will not be much air resistance
• At very large heights, the value of g will also be small
8. Because of all the advantageous conditions, the range achieved will be so large that, the satellite is able to cover the 'curvature of the earth'
• This is shown in fig.8.63 below
9. Even after covering the curvature, the satellite will not separate from the earth because, there is still a 'small gravity' pulling it towards the center of the earth
• Since the air resistance is low, the 'horizontal velocity' remains practically constant
• Since the gravity is low, the height also remains practically constant
• So we have a 'continuous horizontal projectile motion'
• It will take decades or even centuries for the satellite to lower down to the earth's atmosphere
10. From the above steps, it is clear that, the 'motion of a satellite' is a 'horizontal projectile motion'
• Any horizontal projectile is acted upon by gravity
• In the case of satellites however, the gravity is small
• But we cannot ignore it
• The only force acting on the satellite is gravity, which pulls it towards the center of the earth
11. Since the only force acting is gravity, we can say that, the satellite is in free fall
• It is like a lift whose cable is broken
• We saw this case in apparent weight in a lift
• We saw that, a person in such a lift will experience weightlessness
• In the same way, a person in a satellite will also experience weightlessness
Solved example 8.55
Two stars each of one solar mass (= 2 × 1030 kg) are approaching each other for a head on collision. When they are a distance 109 km, their speeds are negligible. What is the speed with which they collide ? The radius of each star is 104 km. Assume the stars to remain undistorted until they collide. (Use the known value of G)
Solution:
1. Consider the point at which the distance between the two stars is 1012 m
• Let this be the initial point
• Given that, the speeds at this point are negligible. So the initial velocity can be considered to be zero
• That means, the initial kinetic energy (Ki) will be zero
2. When the two stars are at a distance of 1012 m, the potential energy (Ui) will be given by:
$\mathbf\small{U_i=-\frac{G\,m\,m}{r}=-\frac{G\,m^2}{10^12}}$
3. So total initial energy = $\mathbf\small{E_i=(U_i+K_i)=(U_i+0)=-\frac{G\,m^2}{10^12}}$
4. Consider the point at which the two stars just collide
• Let this be the final point
• Let v be the velocity of each star at that instant
• Then the total kinetic energy at the final point = $\mathbf\small{K_f=\frac{1}{2}mv^2+\frac{1}{2}mv^2=mv^2}$
5. The distance between the centers of the two stars at the final point = 2 times the radius of each star
= 2 × 107 m
• So potential energy (Uf) at the final point will be given by: $\mathbf\small{U_f=-\frac{G\,m\,m}{r}=-\frac{G\,m^2}{2 \times 10^7}}$
6. Total final energy = $\mathbf\small{E_f=(U_f+K_f)=-\frac{G\,m^2}{2 \times 10^7}+mv^2}$
7. Equating the two energies in (3) and (6), we get: $\mathbf\small{-\frac{G\,m^2}{10^12}=-\frac{G\,m^2}{2 \times 10^7}+mv^2}$
$\mathbf\small{\Rightarrow -\frac{G\,m}{10^12}=-\frac{G\,m}{2 \times 10^7}+v^2}$
$\mathbf\small{\Rightarrow v^2=\frac{G\,m}{2 \times 10^7}-\frac{G\,m}{10^12}}$
$\mathbf\small{\Rightarrow v^2=G\,m \left(\frac{1}{2 \times 10^7}-\frac{1}{10^{12}}\right)=G\,m\times 4.99 \times 10^{-8}}$
$\mathbf\small{\Rightarrow v^2=6.67 \times 10^{-11}\times2\times 10^{30}\times 4.99 \times 10^{-8}}$
$\mathbf\small{\Rightarrow v^2=6.67 \times 10^{12}}$
$\mathbf\small{\Rightarrow v=2.58 \times 10^{6}}$ ms-1
Solved example 8.56
Two heavy spheres each of mass 100 kg and radius 0.10 m are placed 1.0 m apart on a horizontal table. What is the gravitational force and potential at the mid point of the line joining the centres of the spheres ? Is an object placed at that point in equilibrium? If so, is the equilibrium stable or unstable ?
Solution:
Part (i):
1. We can find the 'intensity of gravitational field' at any point due to an object A (Details here)
• We have Eq.8.12: $\mathbf\small{|\vec{E}_{G(A)}|=\frac{G\,\,m_A}{r^2}}$
2. So intensity due to A at the mid point = $\mathbf\small{|\vec{E}_{G(A)}|=\frac{G\,\,m_A}{0.5^2}}$
• This force will be acting towards the left
3. Similarly, intensity due to B at the mid point = $\mathbf\small{|\vec{E}_{G(B)}|=\frac{G\,\,m_B}{0.5^2}}$
• This force will be acting towards the right
4. But both A and B have the same mass of 100 kg
• So the magnitudes of both the forces are equal
• Since they act in opposite directions, the net force at the midpoint will be zero
Part (ii):
1. We can find the 'gravitational potential' at any point due to an object A (Details here)
• We have Eq.8.16: $\mathbf\small{V_A=-\frac{G\,\,m_A}{r}}$
2. So potential due to A at the mid point = $\mathbf\small{V_A=-\frac{G\,\,m_A}{0.5}}$
3. Similarly, potential due to B at the mid point = $\mathbf\small{V_B=-\frac{G\,\,m_B}{0.5}}$
4. So total potential
= $\mathbf\small{V_A+V_B=(-\frac{G\,\,m_A}{0.5}+-\frac{G\,\,m_B}{0.5})=(-\frac{G\,\,m}{0.5}+-\frac{G\,\,m}{0.5})=-4Gm}$
• Substituting the values, we get:
Total potential = (-4 × 6.67 × 10-11 ×100) = 2.67 × 10-8 J kg-1
Part (iii):
• An object C placed at the midpoint will be in equilibrium because, the net force on it will be zero
• But if a small displacement occurs for C, the equilibrium will be lost. It will move towards A or B depending on the direction of displacement
• So the equilibrium of C is an unstable equilibrium
Solved example 8.57
A spaceship is stationed on Mars. How much energy must be expended on the spaceship to launch it out of the solar system ? Mass of the space ship = 1000 kg; mass of the sun = 2 ×1030 kg; mass of mars = 6.4 × 10 23 kg; radius of mars = 3395 km; radius of the orbit of mars = 2.28 × 108 km; G = 6.67 × 10-11 N m2 kg-2.
Solution:
1. When the spaceship is stationed on the surface of mars, it will have a potential energy due to the attraction towards the center of mars
• This potential energy is given by: $\mathbf\small{-\frac{G\,M_M\,m_S}{R_M}}$
• Where
♦ MM is the mass of mars
♦ mS is the mass of the space ship
♦ RM is the radius of mars
2. The spaceship will have an additional potential energy due to the attraction towards the center of sun
• This potential energy is given by: $\mathbf\small{-\frac{G\,M_S\,m_S}{r_S}}$
• Where
♦ MS is the mass of sun
♦ mS is the mass of the space ship
♦ rS is the distance between center of sun and the spaceship
♦ This distance = radius of 'mars orbit' - radius of mars
♦ But radius of mars is negligible when compared to radius of 'mars orbit'
♦ So rS = radius of 'mars orbit'
3. So total potential energy = $\mathbf\small{-\frac{G\,M_M\,m_S}{R_M}-\frac{G\,M_S\,m_S}{r_M}}$
• Substituting the given values, we get:
Potential energy of the spacecraft = -5.98 × 10-11 J
4. So we must expend an energy of 5.98 × 10-11 J to launch the spaceship out of the solar system
Solved example 8.58
A rocket is fired ‘vertically’ from the surface of mars with a speed of 2 km s-1. If 20% of its initial energy is lost due to martian atmospheric resistance, how far will the rocket go from the surface of mars before returning to it ? Mass of mars = 6.4 × 10 23 kg; radius of mars = 3395 km; G = 6.67 × 10-11 N m2 kg-2
Solution:
1. When the spaceship is stationed on the surface of mars, it will have a potential energy due to the attraction towards the center of mars
• This potential energy is given by: $\mathbf\small{-\frac{G\,M_M\,m_S}{R_M}}$
• Where
♦ MM is the mass of mars
♦ mS is the mass of the space ship
♦ RM is the radius of mars
2. Initial kinetiv energy = $\mathbf\small{\frac{1}{2}m_S\,v_i^2}$
3. So total initial energy = $\mathbf\small{-\frac{G\,M_M\,m_S}{R_M}+\frac{1}{2}m_S\,v_i^2}$
= $\mathbf\small{m_S\left(-\frac{G\,M_M}{R_M}+\frac{1}{2}\,v_i^2\right)}$
4. 20% of initial kinetic energy is lost due to martian atmospheric resistance
• So the energy available at the top most point
= initial potential energy + 80% of the initial kinetic energy
= $\mathbf\small{m_S\left(-\frac{G\,M_M}{R_M}+0.8 \times \frac{1}{2}\,v_i^2\right)}$
5. Let h be the maximum height achieved above the surface
• Then, at the top most point, the potential energy will be equal to $\mathbf\small{-\frac{G\,M_M\,m_S}{(R_M+h)}}$
6. At the top most point, the speed will be zero. So the final kinetic energy is zero
7. So the total final energy at the top most point = $\mathbf\small{-\frac{G\,M_M\,m_S}{(R_M+h)}}$
8. Equating the energies in (4) and (7), we get:
$\mathbf\small{m_S\left(-\frac{G\,M_M}{R_M}+0.8 \times \frac{1}{2}\,v_i^2\right)=-\frac{G\,M_M\,m_S}{(R_M+h)}}$
$\mathbf\small{\Rightarrow \left(-\frac{G\,M_M}{R_M}+0.8 \times \frac{1}{2}\,v_i^2\right)=-\frac{G\,M_M}{(R_M+h)}}$
• Substituting the known values, we get:
$\mathbf\small{\left(-\frac{(6.67\times 10^{-11})\,(6.4\times 10^{23})}{(3.395\times 10^{6})}+0.8 \times \frac{1}{2}\,(2000)^2\right)=-\frac{(6.67\times 10^{-11})\,(6.4\times 10^{23})}{(3.395\times 10^{6}+h)}}$
• From this we get: h = 495 km
• In this section we will see Polar satellites
1. In fig.7.59 below, the earth is shown as a blue sphere
 |
| Fig.7.59 |
♦ It passes through the N and S poles
• The direction of rotation is indicated by the orange curved arrow
2. A satellite shown in red color orbits around the earth
• The orbit of the satellite is circular in shape. It is shown in green color
3. We see that, the orbit is vertical
• It passes through the N and S poles
• Such satellites are called Polar satellites
4. Polar satellites have low heights
• The value of h can be in between 500 and 800 km
5. We derived the relation in the previous section: $\mathbf\small{(R_E+h)=\left({\frac{T^2\,G\,M_E}{4\pi^2}}\right)^{1/3}}$
• h and T are the only variables. Both are in the numerators
• So, since the h is low, T will be also be low
6. Since T is low, we can say that:
• The polar satellites require only a lesser time to go around the earth once
• When the earth completes one rotation, the polar satellite would have completed many rotations
7. Consider a camera fixed on the polar satellite
• Since h is low, the camera can capture only a small 'area' of the earth
• This will be clear from fig.7.60 below:
 |
| Fig.8.60 |
• In fig.b, the satellite is at a larger height
(ii) The angle 𝞱, which is the field of view, is same in both the cases
(iii) In fig.a, the camera is able to capture a width of AB
• In fig.b, the camera is able to capture a width of A'B'
(iv) We see that, A'B' is greater than AB
• It is obvious that, when height increases, more details can be captured
• But unfortunately, for polar satellites, the height is low
(v) Since the satellite is going in the vertical direction, the camera will be capturing a 'vertical strip'
• Because of the low h, this strip will be a narrow one
8. In fig.8.61 below, the surface of the earth is divided into strips
 |
| Fig.8.61 |
• So the camera is able to capture the details in the blue strip
(ii) Remember that, the green orbit is fixed. But the earth rotates about the N-S axis
• So various strips come below the orbit
(iii) When the satellite comes for the next round, the next green strip will be below the orbit
• So the camera gets the opportunity to capture the details in the green strip
(iv) This process continues and the camera is able to capture the details in all the strips
9. Remember that, the satellite is at a lower height
• So the pictures captured will have a good resolution
• We get good resolution pictures of the whole earth, strip by strip
• Because of this advantage, polar satellites are used in remote sensing, meteorology, and environmental studies of the earth
Next, we will see some basics about weightlessness
• We saw how to calculate range, time of flight, path of the projectile etc., We will be using those details for our present discussion
1. Imagine person standing on top of a cliff. Let the height of his position above ground be 15 m
• He throws a stone horizontally with a velocity of 10 ms-1
• The stone will land at a horizontal distance (range) of 17.48 m from the foot of the cliff
• The path followed by the stone is the curve A in fig.8.62 below:
 |
| Fig.8.62 |
• He throws a stone horizontally with the same velocity of 10 ms-1
• This time the range is 22.58 m
• The path followed by the stone is the curve B in fig.8.62
3. Let the person be at the same height of 25 m
• He throws a stone horizontally with an increased velocity of 15 ms-1
• This time the range is 33.86 m
• The path followed by the stone is the curve C
4. Let the person climb much higher to 40 m
• He throws a stone horizontally with a high velocity of 35 ms-1
• This time the range is 99.95 m
• The path followed by the stone is the curve D
5. From the above 4 steps, we can write the following 2 points:
(i) Height and velocity are the only variables possible
• In other words, we can 'make adjustments' only on those two items. Other item 'g' is a constant
(ii) When height and velocity are increased, we can achieve a large range
• Note that, we have neglected the effect of air resistance in the calculations
6. Consider the point 5 (ii)
• We need to increase height and velocity
• This is exactly what we do in the case of a satellite
(i) The rocket carries the satellite to a very large height
(ii) At that height, small rockets are fired so that the satellite separates away from the main rocket
• The satellite separates away in a horizontal direction
(iii) So the motion of the satellite will be similar to the projectile motions that we saw in fig.8.62 above
7. This motion of the satellite has some advantages:
• Since height and velocity are large, the range will be very large
• At very large heights, the air will be so thin that, there will not be much air resistance
• At very large heights, the value of g will also be small
8. Because of all the advantageous conditions, the range achieved will be so large that, the satellite is able to cover the 'curvature of the earth'
• This is shown in fig.8.63 below
 |
| Fig.8.63 |
• Since the air resistance is low, the 'horizontal velocity' remains practically constant
• Since the gravity is low, the height also remains practically constant
• So we have a 'continuous horizontal projectile motion'
• It will take decades or even centuries for the satellite to lower down to the earth's atmosphere
10. From the above steps, it is clear that, the 'motion of a satellite' is a 'horizontal projectile motion'
• Any horizontal projectile is acted upon by gravity
• In the case of satellites however, the gravity is small
• But we cannot ignore it
• The only force acting on the satellite is gravity, which pulls it towards the center of the earth
11. Since the only force acting is gravity, we can say that, the satellite is in free fall
• It is like a lift whose cable is broken
• We saw this case in apparent weight in a lift
• We saw that, a person in such a lift will experience weightlessness
• In the same way, a person in a satellite will also experience weightlessness
Now we will see some solved examples
Solved example 8.55
Two stars each of one solar mass (= 2 × 1030 kg) are approaching each other for a head on collision. When they are a distance 109 km, their speeds are negligible. What is the speed with which they collide ? The radius of each star is 104 km. Assume the stars to remain undistorted until they collide. (Use the known value of G)
Solution:
1. Consider the point at which the distance between the two stars is 1012 m
• Let this be the initial point
• Given that, the speeds at this point are negligible. So the initial velocity can be considered to be zero
• That means, the initial kinetic energy (Ki) will be zero
2. When the two stars are at a distance of 1012 m, the potential energy (Ui) will be given by:
$\mathbf\small{U_i=-\frac{G\,m\,m}{r}=-\frac{G\,m^2}{10^12}}$
3. So total initial energy = $\mathbf\small{E_i=(U_i+K_i)=(U_i+0)=-\frac{G\,m^2}{10^12}}$
4. Consider the point at which the two stars just collide
• Let this be the final point
• Let v be the velocity of each star at that instant
• Then the total kinetic energy at the final point = $\mathbf\small{K_f=\frac{1}{2}mv^2+\frac{1}{2}mv^2=mv^2}$
5. The distance between the centers of the two stars at the final point = 2 times the radius of each star
= 2 × 107 m
• So potential energy (Uf) at the final point will be given by: $\mathbf\small{U_f=-\frac{G\,m\,m}{r}=-\frac{G\,m^2}{2 \times 10^7}}$
6. Total final energy = $\mathbf\small{E_f=(U_f+K_f)=-\frac{G\,m^2}{2 \times 10^7}+mv^2}$
7. Equating the two energies in (3) and (6), we get: $\mathbf\small{-\frac{G\,m^2}{10^12}=-\frac{G\,m^2}{2 \times 10^7}+mv^2}$
$\mathbf\small{\Rightarrow -\frac{G\,m}{10^12}=-\frac{G\,m}{2 \times 10^7}+v^2}$
$\mathbf\small{\Rightarrow v^2=\frac{G\,m}{2 \times 10^7}-\frac{G\,m}{10^12}}$
$\mathbf\small{\Rightarrow v^2=G\,m \left(\frac{1}{2 \times 10^7}-\frac{1}{10^{12}}\right)=G\,m\times 4.99 \times 10^{-8}}$
$\mathbf\small{\Rightarrow v^2=6.67 \times 10^{-11}\times2\times 10^{30}\times 4.99 \times 10^{-8}}$
$\mathbf\small{\Rightarrow v^2=6.67 \times 10^{12}}$
$\mathbf\small{\Rightarrow v=2.58 \times 10^{6}}$ ms-1
Solved example 8.56
Two heavy spheres each of mass 100 kg and radius 0.10 m are placed 1.0 m apart on a horizontal table. What is the gravitational force and potential at the mid point of the line joining the centres of the spheres ? Is an object placed at that point in equilibrium? If so, is the equilibrium stable or unstable ?
Solution:
Part (i):
1. We can find the 'intensity of gravitational field' at any point due to an object A (Details here)
• We have Eq.8.12: $\mathbf\small{|\vec{E}_{G(A)}|=\frac{G\,\,m_A}{r^2}}$
2. So intensity due to A at the mid point = $\mathbf\small{|\vec{E}_{G(A)}|=\frac{G\,\,m_A}{0.5^2}}$
• This force will be acting towards the left
3. Similarly, intensity due to B at the mid point = $\mathbf\small{|\vec{E}_{G(B)}|=\frac{G\,\,m_B}{0.5^2}}$
• This force will be acting towards the right
4. But both A and B have the same mass of 100 kg
• So the magnitudes of both the forces are equal
• Since they act in opposite directions, the net force at the midpoint will be zero
Part (ii):
1. We can find the 'gravitational potential' at any point due to an object A (Details here)
• We have Eq.8.16: $\mathbf\small{V_A=-\frac{G\,\,m_A}{r}}$
2. So potential due to A at the mid point = $\mathbf\small{V_A=-\frac{G\,\,m_A}{0.5}}$
3. Similarly, potential due to B at the mid point = $\mathbf\small{V_B=-\frac{G\,\,m_B}{0.5}}$
4. So total potential
= $\mathbf\small{V_A+V_B=(-\frac{G\,\,m_A}{0.5}+-\frac{G\,\,m_B}{0.5})=(-\frac{G\,\,m}{0.5}+-\frac{G\,\,m}{0.5})=-4Gm}$
• Substituting the values, we get:
Total potential = (-4 × 6.67 × 10-11 ×100) = 2.67 × 10-8 J kg-1
Part (iii):
• An object C placed at the midpoint will be in equilibrium because, the net force on it will be zero
• But if a small displacement occurs for C, the equilibrium will be lost. It will move towards A or B depending on the direction of displacement
• So the equilibrium of C is an unstable equilibrium
Solved example 8.57
A spaceship is stationed on Mars. How much energy must be expended on the spaceship to launch it out of the solar system ? Mass of the space ship = 1000 kg; mass of the sun = 2 ×1030 kg; mass of mars = 6.4 × 10 23 kg; radius of mars = 3395 km; radius of the orbit of mars = 2.28 × 108 km; G = 6.67 × 10-11 N m2 kg-2.
Solution:
1. When the spaceship is stationed on the surface of mars, it will have a potential energy due to the attraction towards the center of mars
• This potential energy is given by: $\mathbf\small{-\frac{G\,M_M\,m_S}{R_M}}$
• Where
♦ MM is the mass of mars
♦ mS is the mass of the space ship
♦ RM is the radius of mars
2. The spaceship will have an additional potential energy due to the attraction towards the center of sun
• This potential energy is given by: $\mathbf\small{-\frac{G\,M_S\,m_S}{r_S}}$
• Where
♦ MS is the mass of sun
♦ mS is the mass of the space ship
♦ rS is the distance between center of sun and the spaceship
♦ This distance = radius of 'mars orbit' - radius of mars
♦ But radius of mars is negligible when compared to radius of 'mars orbit'
♦ So rS = radius of 'mars orbit'
3. So total potential energy = $\mathbf\small{-\frac{G\,M_M\,m_S}{R_M}-\frac{G\,M_S\,m_S}{r_M}}$
• Substituting the given values, we get:
Potential energy of the spacecraft = -5.98 × 10-11 J
4. So we must expend an energy of 5.98 × 10-11 J to launch the spaceship out of the solar system
Solved example 8.58
A rocket is fired ‘vertically’ from the surface of mars with a speed of 2 km s-1. If 20% of its initial energy is lost due to martian atmospheric resistance, how far will the rocket go from the surface of mars before returning to it ? Mass of mars = 6.4 × 10 23 kg; radius of mars = 3395 km; G = 6.67 × 10-11 N m2 kg-2
Solution:
1. When the spaceship is stationed on the surface of mars, it will have a potential energy due to the attraction towards the center of mars
• This potential energy is given by: $\mathbf\small{-\frac{G\,M_M\,m_S}{R_M}}$
• Where
♦ MM is the mass of mars
♦ mS is the mass of the space ship
♦ RM is the radius of mars
2. Initial kinetiv energy = $\mathbf\small{\frac{1}{2}m_S\,v_i^2}$
3. So total initial energy = $\mathbf\small{-\frac{G\,M_M\,m_S}{R_M}+\frac{1}{2}m_S\,v_i^2}$
= $\mathbf\small{m_S\left(-\frac{G\,M_M}{R_M}+\frac{1}{2}\,v_i^2\right)}$
4. 20% of initial kinetic energy is lost due to martian atmospheric resistance
• So the energy available at the top most point
= initial potential energy + 80% of the initial kinetic energy
= $\mathbf\small{m_S\left(-\frac{G\,M_M}{R_M}+0.8 \times \frac{1}{2}\,v_i^2\right)}$
5. Let h be the maximum height achieved above the surface
• Then, at the top most point, the potential energy will be equal to $\mathbf\small{-\frac{G\,M_M\,m_S}{(R_M+h)}}$
6. At the top most point, the speed will be zero. So the final kinetic energy is zero
7. So the total final energy at the top most point = $\mathbf\small{-\frac{G\,M_M\,m_S}{(R_M+h)}}$
8. Equating the energies in (4) and (7), we get:
$\mathbf\small{m_S\left(-\frac{G\,M_M}{R_M}+0.8 \times \frac{1}{2}\,v_i^2\right)=-\frac{G\,M_M\,m_S}{(R_M+h)}}$
$\mathbf\small{\Rightarrow \left(-\frac{G\,M_M}{R_M}+0.8 \times \frac{1}{2}\,v_i^2\right)=-\frac{G\,M_M}{(R_M+h)}}$
• Substituting the known values, we get:
$\mathbf\small{\left(-\frac{(6.67\times 10^{-11})\,(6.4\times 10^{23})}{(3.395\times 10^{6})}+0.8 \times \frac{1}{2}\,(2000)^2\right)=-\frac{(6.67\times 10^{-11})\,(6.4\times 10^{23})}{(3.395\times 10^{6}+h)}}$
• From this we get: h = 495 km
So we have completed our present discussion on Gravitation. In the next chapter we will see Mechanical properties of solids
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