In the previous section, we completed a discussion on escape velocity
• In this section we will see Earth satellites
First we will find the speed with which satellites move around the earth
1. Consider a satellite moving around the earth in a circular orbit
• Let it's mass be m
• Let it's speed be V
• Let it be at a height h above the surface of the earth
♦ So the radius r of the circular orbit will be equal to (RE + h)
2. Any object moving in a circular path requires centripetal force
• We know that, for our present case, the centripetal force will be equal to $\mathbf\small{\frac{mV^2}{R_E+h}}$
3. This centripetal force is provided by the gravitational force between the earth and the satellite
• We know that, the gravitational force will be equal to $\mathbf\small{\frac{G\,M_E\,m}{(R_E+h)^2}}$
4. Equating the results in (2) and (3), we get: $\mathbf\small{\frac{mV^2}{R_E+h}=\frac{G\,M_E\,m}{(R_E+h)^2}}$
$\mathbf\small{\Rightarrow \frac{V^2}{R_E+h}=\frac{G\,M_E}{(R_E+h)^2}}$
$\mathbf\small{\Rightarrow V^2=\frac{G\,M_E}{(R_E+h)}}$
Thus we get:
Eq.8.22: $\mathbf\small{V=\sqrt{\frac{G\,M_E}{(R_E+h)}}}$
5. In the above equation, there is no m
• All quantities except h are constants
• The h is in the denominator
• So we can write:
♦ The speed of a satellite does not depend on it's mass
♦ When h increases, speed decreases
6. If the satellite is very close to the surface of the earth, (RE+h) can be taken approximately equal to RE
• In such cases, we can write a separate equation:
Speed of satellites very close to the surface of the earth is given by:
Eq.8.23: $\mathbf\small{V=\sqrt{\frac{G\,M_E}{R_E}}}$
7. In the above equation 8.23, let us multiply both numerator and denominator by ✓RE
• We get: $\mathbf\small{V=\sqrt{\frac{G\,M_E\,R_E}{R_E^2}}}$
• Thus we get:
Speed of satellites very close to the surface of the earth is given by:
Eq.8.24: $\mathbf\small{V=\sqrt{g\,R_E}}$
(∵ $\mathbf\small{g={\frac{G\,M_E}{R_E^2}}}$)
1. Let the time period of an earth satellite be T
• That means., T seconds are required by that satellite to complete one rotation around the earth
2. Obviously, during those T seconds, the satellite will travel a distance equal to the 'circumference of it's orbit'
• This circumference is equal to $\mathbf\small{2\pi(R_E+h)}$
3. When we divide 'distance traveled' by time, we get the speed
• So we can write: $\mathbf\small{V={\frac{2\pi(R_E+h)}{T}}}$
4. But we have already calculated V
• Putting that value, the result in (3) becomes:
$\mathbf\small{\sqrt{\frac{G\,M_E}{(R_E+h)}}={\frac{2\pi(R_E+h)}{T}}}$
• Squaring both sides, we get: $\mathbf\small{\frac{G\,M_E}{(R_E+h)}={\frac{4\pi^2(R_E+h)^2}{T^2}}}$
$\mathbf\small{\Rightarrow T^2={\frac{4\pi^2(R_E+h)^3}{G\,M_E}}}$
$\mathbf\small{\Rightarrow T^2={\frac{4\pi^2}{G\,M_E}}(R_E+h)^3}$
5. $\mathbf\small{\frac{4\pi^2}{G\,M_E}}$ is a constant. So we can write:
Eq.8.25: $\mathbf\small{T^2=k(R_E+h)^3}$
Where k = $\mathbf\small{\frac{4\pi^2}{G\,M_E}}$ = a constant
• So we can write:
The square of the 'time period of an earth satellite' is proportional to the cube of the 'distance of that planet from the center of the earth'
• Thus it is clear that, earth satellites obey Kepler's third law
6. From the result in (4), we can obtain an expression for the time period:
Eq.8.26: $\mathbf\small{T={\frac{2\pi(R_E+h)^{3/2}}{\sqrt{G\,M_E}}}}$
7. If the satellite is very close to the surface of the earth, (RE+h) can be taken approximately equal to RE
• Then we can rearrange Eq.8.26:
$\mathbf\small{T={\frac{2\pi(R_E)^{3/2}}{\sqrt{G\,M_E}}}}$
• Squaring both sides, we get: $\mathbf\small{T^2={\frac{4\pi^2(R_E)^{3}}{G\,M_E}}}$
$\mathbf\small{\Rightarrow T^2=4\pi^2 \left(\frac{R_E^{2}}{G\,M_E}\right)R_E}$
• But $\mathbf\small{\left(\frac{R_E^{2}}{G\,M_E}\right)}$ is $\mathbf\small{\frac{1}{g}}$
• So we get: $\mathbf\small{T^2=4\pi^2\frac{R_E}{g}}$
• Thus we get:
Time period of satellites very close to the surface of the earth is given by:
Eq.8.27: $\mathbf\small{T=2\pi\sqrt{\frac{R_E}{g}}}$
8. Let us put the known values in Eq.8.26. We get: $\mathbf\small{T=2\pi\sqrt{\frac{6.4\times 10^6}{9.8}}}$
• This works out to approximately 85 minutes
• So we can write:
Satellites which are close to the earth will have a time period of approximately 85 minutes
Solved example 8.43
An artificial satellite very close to the surface of the earth, revolves with a speed v. What will be the speed of another artificial satellite, whose height from the surface is 0.5RE ?
Solution:
1. Given that, the satellite is very close to the surface of the earth
• So we can use Eq.8.23: $\mathbf\small{V=\sqrt{\frac{G\,M_E}{R_E}}}$
• Substituting the given value 'v', we get: $\mathbf\small{v=\sqrt{\frac{G\,M_E}{R_E}}}$
2. Now we want the speed of another satellite whose h is 0.5RE
• We can use Eq.8.22: $\mathbf\small{V=\sqrt{\frac{G\,M_E}{(R_E+h)}}}$
• Let v' be the speed of this satellite
• Substituting the values, we get: $\mathbf\small{v'=\sqrt{\frac{G\,M_E}{(1.5R_E)}}}$
3. Taking ratios, we get:
$\mathbf\small{\frac{v}{v'}=\sqrt{\frac{G\,M_E}{R_E}}\times \sqrt{\frac{1.5R_E}{G\,M_E}}=\sqrt{1.5}}$
• Thus we get: $\mathbf\small{v'=\frac{v}{\sqrt{1.5}}}$
Solved example 8.44
Two satellites A and B revolve around a planet in orbits of radii 4R and R respectively. If the speed of the satellite A is 3v, what is the speed of B?
Solution:
1. We can use Eq.8.22: $\mathbf\small{V=\sqrt{\frac{G\,M_E}{(R_E+h)}}}$
• Substituting the values we get:
$\mathbf\small{V_A=\sqrt{\frac{G\,M_E}{4R}}}$
$\mathbf\small{V_B=\sqrt{\frac{G\,M_E}{R}}}$
2. Taking ratios, we get:
$\mathbf\small{\frac{V_A}{V_B}=\sqrt{\frac{G\,M_E}{4R}}\times \sqrt{\frac{R}{G\,M_E}}=\sqrt{\frac{1}{4}}=\frac{1}{2}}$
3. But given that VA = 3v
• So we get: $\mathbf\small{V_B=\sqrt{2}\,V_A=3\times 2\,v=6v}$
Solved example 8.45
The radii of the orbits of two satellites A and B are in the ratio 1:4. Calculate TA : TB
Solution:
1. We can use Kepler's law:
Eq.8.25: $\mathbf\small{T^2=k(R_E+h)^3}$
$\mathbf\small{\Rightarrow T^2=k\;r^3}$
• Where (RE+h) = r = distance from the center of the planet = radius of the orbit
2. Substituting the values, we get:
$\mathbf\small{T_A^2=k\;r_A^3}$
$\mathbf\small{T_B^2=k\;r_B^3}$
3. Taking ratios. we get:
$\mathbf\small{\frac{T_A^2}{T_B^2}=\frac{r_A^3}{r_B^3}\Rightarrow \left(\frac{T_A}{T_B}\right)^2=\left(\frac{r_A}{r_B}\right)^3}$
$\mathbf\small{\Rightarrow \left(\frac{T_A}{T_B}\right)^2=\left(\frac{1}{4}\right)^3=\frac{1}{64}}$
$\mathbf\small{\Rightarrow \frac{T_A}{T_B}=\frac{1}{8}}$
Solved example 8.46
The planet Mars has two moons, Phobos and Delmos. (i) Phobos has a period 7 hours, 39 minutes and an orbital radius of 9.4 × 103 km. Calculate the mass of mars. (ii) Assume that earth and mars move in circular orbits around the sun, with the martian orbit being 1.52 times the orbital radius of the earth. What is the length of the martian year in days ?
Solution:
Part (i):
1. We are given the time period. So we will use an equation connecting T and mass
• We have Eq.8.26 for an earth satellite: $\mathbf\small{T={\frac{2\pi(R_E+h)^{3/2}}{\sqrt{G\,M_E}}}}$
2. For a Mars satellite, we can write: $\mathbf\small{T={\frac{2\pi(R_M+h)^{3/2}}{\sqrt{G\,M_M}}}}$
• Substituting the values, we get:
$\mathbf\small{\left[(7)(60)+39\right](60)={\frac{2\pi\left[(9.4)(10)^3 (10)^3\right]^{3/2}}{\sqrt{(6.67)(10^{-11})\,M_M}}}}$
3. The mass is the only unknown quantity. So we get: MM = 6.48 × 1023 kg
Part(ii):
1. We can use Kepler's law:
Eq.8.25: $\mathbf\small{T^2=k(R_E+h)^3}$
$\mathbf\small{\Rightarrow T^2=k\;r^3}$
• Where (RE+h) = r = distance from the center of the planet = radius of the orbit
2. Substituting the values, we get:
$\mathbf\small{T_E^2=k\;r_E^3}$
$\mathbf\small{T_M^2=k\;r_M^3}$
3. Taking ratios. we get:
$\mathbf\small{\frac{T_E^2}{T_M^2}=\frac{r_E^3}{r_M^3}\Rightarrow \left(\frac{T_E}{T_M}\right)^2=\left(\frac{r_E}{r_M}\right)^3}$
$\mathbf\small{\Rightarrow \left(\frac{T_E}{T_M}\right)^2=\left(\frac{r_E}{1.52\,r_E}\right)^3=\frac{1}{1.52^3}}$
4. But TE = 365 days. So we get:
$\mathbf\small{\frac{365^2}{T_M^2}=\frac{1}{1.52^3}}$
• Thus we get: TM = 684 days
Solved example 8.47
You are given the following data: g = 9.81 m s-2, RE = 6.37 × 106 m, the distance to the moon R = 3.84 × 108 m and the time period of the moon’s revolution is 27.3 days. Obtain the mass of the Earth ME in two different ways.
Solution:
Method 1:
We will use an equation which connects mass and force
1. Consider a body of mass m resting on the surface of the earth
• The gravitational force of attraction acting on it towards the center of the earth is $\mathbf\small{\frac{G\,M_E\,m}{R_E^2}}$
2. But this force is the weight mg of the body
3. Equating the two, we get: $\mathbf\small{mg=\frac{G\,M_E\,m}{R_E^2}}$
$\mathbf\small{\Rightarrow g=\frac{G\,M_E}{R_E^2}}$
4. Substituting the known values, we get: $\mathbf\small{9.81=\frac{(6.67 \times 10^{-11})\,M_E}{(6.37 \times 10^{6})^2}}$
• ME is the only unknown quantity. So we get:
ME = 5.97 × 1024 kg
Method 2:
• We will use an equation which connects mass and time period T
• We have Eq.8.26: $\mathbf\small{T={\frac{2\pi(R_E+h)^{3/2}}{\sqrt{G\,M_E}}}}$
• Substituting the values, we get: $\mathbf\small{(27.3)(24\times 60\times 60)={\frac{2\pi(3.84\times 10^8)^{3/2}}{\sqrt{(6.67\times 10^{-11})\,M_E}}}}$
• ME is the only unknown quantity. So we get:
ME = 6.024 × 1024 kg
■ The mass obtained by the two methods are approximately equal
Solved example 8.48
Express the constant k of Eq. (8.25) in days and km. Given k = 10-13 s2 m-3. The moon is at a distance of 3.84 × 105 km from the earth. Obtain its time-period of revolution in days.
Solution:
Part (i):
1. We have Eq.8.25: $\mathbf\small{T^2=k(R_E+h)^3}$
• Where k = $\mathbf\small{\frac{4\pi^2}{G\,M_E}}$ = a constant
2. The equation can be rearranged as $\mathbf\small{k=\frac{T^2}{(R_E+h)^3}}$
3. In SI system, the unit of time is s and the unit of distance is m
• So the units of k can be calculated as: $\mathbf\small{k=\frac{s^2}{m^3}}$
4. We want time in 'terms of days' and distance in 'terms of km'
• 1 s = $\mathbf\small{\frac{1}{24 \times 60 \times 60}=\frac{1}{86400}}$ days
• 1 m = 10-3 km
5. So 1 s2 m-3 = $\mathbf\small{\frac{(\frac{1}{86400})^2}{(10^{-3})^3}}$
• So 10-13 s2 m-3 = $\mathbf\small{10^{-13}\times \frac{(\frac{1}{86400})^2}{(10^{-3})^3}}$ = 1.33 × 10-14 days2 km-3
Part (ii):
• Using Eq.8.25, we get: $\mathbf\small{1.33 \times 10^{-14}\times 3.84 \times 10^5}$ = 753.08
• Thus T = ✓(753.08) = 27.3 days
• In this section we will see Earth satellites
First we will find the speed with which satellites move around the earth
1. Consider a satellite moving around the earth in a circular orbit
• Let it's mass be m
• Let it's speed be V
• Let it be at a height h above the surface of the earth
♦ So the radius r of the circular orbit will be equal to (RE + h)
2. Any object moving in a circular path requires centripetal force
• We know that, for our present case, the centripetal force will be equal to $\mathbf\small{\frac{mV^2}{R_E+h}}$
3. This centripetal force is provided by the gravitational force between the earth and the satellite
• We know that, the gravitational force will be equal to $\mathbf\small{\frac{G\,M_E\,m}{(R_E+h)^2}}$
4. Equating the results in (2) and (3), we get: $\mathbf\small{\frac{mV^2}{R_E+h}=\frac{G\,M_E\,m}{(R_E+h)^2}}$
$\mathbf\small{\Rightarrow \frac{V^2}{R_E+h}=\frac{G\,M_E}{(R_E+h)^2}}$
$\mathbf\small{\Rightarrow V^2=\frac{G\,M_E}{(R_E+h)}}$
Thus we get:
Eq.8.22: $\mathbf\small{V=\sqrt{\frac{G\,M_E}{(R_E+h)}}}$
5. In the above equation, there is no m
• All quantities except h are constants
• The h is in the denominator
• So we can write:
♦ The speed of a satellite does not depend on it's mass
♦ When h increases, speed decreases
6. If the satellite is very close to the surface of the earth, (RE+h) can be taken approximately equal to RE
• In such cases, we can write a separate equation:
Speed of satellites very close to the surface of the earth is given by:
Eq.8.23: $\mathbf\small{V=\sqrt{\frac{G\,M_E}{R_E}}}$
7. In the above equation 8.23, let us multiply both numerator and denominator by ✓RE
• We get: $\mathbf\small{V=\sqrt{\frac{G\,M_E\,R_E}{R_E^2}}}$
• Thus we get:
Speed of satellites very close to the surface of the earth is given by:
Eq.8.24: $\mathbf\small{V=\sqrt{g\,R_E}}$
(∵ $\mathbf\small{g={\frac{G\,M_E}{R_E^2}}}$)
Next we want the time period T of a satellite
• That means., T seconds are required by that satellite to complete one rotation around the earth
2. Obviously, during those T seconds, the satellite will travel a distance equal to the 'circumference of it's orbit'
• This circumference is equal to $\mathbf\small{2\pi(R_E+h)}$
3. When we divide 'distance traveled' by time, we get the speed
• So we can write: $\mathbf\small{V={\frac{2\pi(R_E+h)}{T}}}$
4. But we have already calculated V
• Putting that value, the result in (3) becomes:
$\mathbf\small{\sqrt{\frac{G\,M_E}{(R_E+h)}}={\frac{2\pi(R_E+h)}{T}}}$
• Squaring both sides, we get: $\mathbf\small{\frac{G\,M_E}{(R_E+h)}={\frac{4\pi^2(R_E+h)^2}{T^2}}}$
$\mathbf\small{\Rightarrow T^2={\frac{4\pi^2(R_E+h)^3}{G\,M_E}}}$
$\mathbf\small{\Rightarrow T^2={\frac{4\pi^2}{G\,M_E}}(R_E+h)^3}$
5. $\mathbf\small{\frac{4\pi^2}{G\,M_E}}$ is a constant. So we can write:
Eq.8.25: $\mathbf\small{T^2=k(R_E+h)^3}$
Where k = $\mathbf\small{\frac{4\pi^2}{G\,M_E}}$ = a constant
• So we can write:
The square of the 'time period of an earth satellite' is proportional to the cube of the 'distance of that planet from the center of the earth'
• Thus it is clear that, earth satellites obey Kepler's third law
6. From the result in (4), we can obtain an expression for the time period:
Eq.8.26: $\mathbf\small{T={\frac{2\pi(R_E+h)^{3/2}}{\sqrt{G\,M_E}}}}$
7. If the satellite is very close to the surface of the earth, (RE+h) can be taken approximately equal to RE
• Then we can rearrange Eq.8.26:
$\mathbf\small{T={\frac{2\pi(R_E)^{3/2}}{\sqrt{G\,M_E}}}}$
• Squaring both sides, we get: $\mathbf\small{T^2={\frac{4\pi^2(R_E)^{3}}{G\,M_E}}}$
$\mathbf\small{\Rightarrow T^2=4\pi^2 \left(\frac{R_E^{2}}{G\,M_E}\right)R_E}$
• But $\mathbf\small{\left(\frac{R_E^{2}}{G\,M_E}\right)}$ is $\mathbf\small{\frac{1}{g}}$
• So we get: $\mathbf\small{T^2=4\pi^2\frac{R_E}{g}}$
• Thus we get:
Time period of satellites very close to the surface of the earth is given by:
Eq.8.27: $\mathbf\small{T=2\pi\sqrt{\frac{R_E}{g}}}$
8. Let us put the known values in Eq.8.26. We get: $\mathbf\small{T=2\pi\sqrt{\frac{6.4\times 10^6}{9.8}}}$
• This works out to approximately 85 minutes
• So we can write:
Satellites which are close to the earth will have a time period of approximately 85 minutes
So we have seen speed (V) and time period (T). Many other properties of celestial bodies can be calculated based on these two items. Some solved examples given below will demonstrate this concept:
Solved example 8.43
An artificial satellite very close to the surface of the earth, revolves with a speed v. What will be the speed of another artificial satellite, whose height from the surface is 0.5RE ?
Solution:
1. Given that, the satellite is very close to the surface of the earth
• So we can use Eq.8.23: $\mathbf\small{V=\sqrt{\frac{G\,M_E}{R_E}}}$
• Substituting the given value 'v', we get: $\mathbf\small{v=\sqrt{\frac{G\,M_E}{R_E}}}$
2. Now we want the speed of another satellite whose h is 0.5RE
• We can use Eq.8.22: $\mathbf\small{V=\sqrt{\frac{G\,M_E}{(R_E+h)}}}$
• Let v' be the speed of this satellite
• Substituting the values, we get: $\mathbf\small{v'=\sqrt{\frac{G\,M_E}{(1.5R_E)}}}$
3. Taking ratios, we get:
$\mathbf\small{\frac{v}{v'}=\sqrt{\frac{G\,M_E}{R_E}}\times \sqrt{\frac{1.5R_E}{G\,M_E}}=\sqrt{1.5}}$
• Thus we get: $\mathbf\small{v'=\frac{v}{\sqrt{1.5}}}$
Solved example 8.44
Two satellites A and B revolve around a planet in orbits of radii 4R and R respectively. If the speed of the satellite A is 3v, what is the speed of B?
Solution:
1. We can use Eq.8.22: $\mathbf\small{V=\sqrt{\frac{G\,M_E}{(R_E+h)}}}$
• Substituting the values we get:
$\mathbf\small{V_A=\sqrt{\frac{G\,M_E}{4R}}}$
$\mathbf\small{V_B=\sqrt{\frac{G\,M_E}{R}}}$
2. Taking ratios, we get:
$\mathbf\small{\frac{V_A}{V_B}=\sqrt{\frac{G\,M_E}{4R}}\times \sqrt{\frac{R}{G\,M_E}}=\sqrt{\frac{1}{4}}=\frac{1}{2}}$
3. But given that VA = 3v
• So we get: $\mathbf\small{V_B=\sqrt{2}\,V_A=3\times 2\,v=6v}$
Solved example 8.45
The radii of the orbits of two satellites A and B are in the ratio 1:4. Calculate TA : TB
Solution:
1. We can use Kepler's law:
Eq.8.25: $\mathbf\small{T^2=k(R_E+h)^3}$
$\mathbf\small{\Rightarrow T^2=k\;r^3}$
• Where (RE+h) = r = distance from the center of the planet = radius of the orbit
2. Substituting the values, we get:
$\mathbf\small{T_A^2=k\;r_A^3}$
$\mathbf\small{T_B^2=k\;r_B^3}$
3. Taking ratios. we get:
$\mathbf\small{\frac{T_A^2}{T_B^2}=\frac{r_A^3}{r_B^3}\Rightarrow \left(\frac{T_A}{T_B}\right)^2=\left(\frac{r_A}{r_B}\right)^3}$
$\mathbf\small{\Rightarrow \left(\frac{T_A}{T_B}\right)^2=\left(\frac{1}{4}\right)^3=\frac{1}{64}}$
$\mathbf\small{\Rightarrow \frac{T_A}{T_B}=\frac{1}{8}}$
Solved example 8.46
The planet Mars has two moons, Phobos and Delmos. (i) Phobos has a period 7 hours, 39 minutes and an orbital radius of 9.4 × 103 km. Calculate the mass of mars. (ii) Assume that earth and mars move in circular orbits around the sun, with the martian orbit being 1.52 times the orbital radius of the earth. What is the length of the martian year in days ?
Solution:
Part (i):
1. We are given the time period. So we will use an equation connecting T and mass
• We have Eq.8.26 for an earth satellite: $\mathbf\small{T={\frac{2\pi(R_E+h)^{3/2}}{\sqrt{G\,M_E}}}}$
2. For a Mars satellite, we can write: $\mathbf\small{T={\frac{2\pi(R_M+h)^{3/2}}{\sqrt{G\,M_M}}}}$
• Substituting the values, we get:
$\mathbf\small{\left[(7)(60)+39\right](60)={\frac{2\pi\left[(9.4)(10)^3 (10)^3\right]^{3/2}}{\sqrt{(6.67)(10^{-11})\,M_M}}}}$
3. The mass is the only unknown quantity. So we get: MM = 6.48 × 1023 kg
Part(ii):
1. We can use Kepler's law:
Eq.8.25: $\mathbf\small{T^2=k(R_E+h)^3}$
$\mathbf\small{\Rightarrow T^2=k\;r^3}$
• Where (RE+h) = r = distance from the center of the planet = radius of the orbit
2. Substituting the values, we get:
$\mathbf\small{T_E^2=k\;r_E^3}$
$\mathbf\small{T_M^2=k\;r_M^3}$
3. Taking ratios. we get:
$\mathbf\small{\frac{T_E^2}{T_M^2}=\frac{r_E^3}{r_M^3}\Rightarrow \left(\frac{T_E}{T_M}\right)^2=\left(\frac{r_E}{r_M}\right)^3}$
$\mathbf\small{\Rightarrow \left(\frac{T_E}{T_M}\right)^2=\left(\frac{r_E}{1.52\,r_E}\right)^3=\frac{1}{1.52^3}}$
4. But TE = 365 days. So we get:
$\mathbf\small{\frac{365^2}{T_M^2}=\frac{1}{1.52^3}}$
• Thus we get: TM = 684 days
Solved example 8.47
You are given the following data: g = 9.81 m s-2, RE = 6.37 × 106 m, the distance to the moon R = 3.84 × 108 m and the time period of the moon’s revolution is 27.3 days. Obtain the mass of the Earth ME in two different ways.
Solution:
Method 1:
We will use an equation which connects mass and force
1. Consider a body of mass m resting on the surface of the earth
• The gravitational force of attraction acting on it towards the center of the earth is $\mathbf\small{\frac{G\,M_E\,m}{R_E^2}}$
2. But this force is the weight mg of the body
3. Equating the two, we get: $\mathbf\small{mg=\frac{G\,M_E\,m}{R_E^2}}$
$\mathbf\small{\Rightarrow g=\frac{G\,M_E}{R_E^2}}$
4. Substituting the known values, we get: $\mathbf\small{9.81=\frac{(6.67 \times 10^{-11})\,M_E}{(6.37 \times 10^{6})^2}}$
• ME is the only unknown quantity. So we get:
ME = 5.97 × 1024 kg
Method 2:
• We will use an equation which connects mass and time period T
• We have Eq.8.26: $\mathbf\small{T={\frac{2\pi(R_E+h)^{3/2}}{\sqrt{G\,M_E}}}}$
• Substituting the values, we get: $\mathbf\small{(27.3)(24\times 60\times 60)={\frac{2\pi(3.84\times 10^8)^{3/2}}{\sqrt{(6.67\times 10^{-11})\,M_E}}}}$
• ME is the only unknown quantity. So we get:
ME = 6.024 × 1024 kg
■ The mass obtained by the two methods are approximately equal
Solved example 8.48
Express the constant k of Eq. (8.25) in days and km. Given k = 10-13 s2 m-3. The moon is at a distance of 3.84 × 105 km from the earth. Obtain its time-period of revolution in days.
Solution:
Part (i):
1. We have Eq.8.25: $\mathbf\small{T^2=k(R_E+h)^3}$
• Where k = $\mathbf\small{\frac{4\pi^2}{G\,M_E}}$ = a constant
2. The equation can be rearranged as $\mathbf\small{k=\frac{T^2}{(R_E+h)^3}}$
3. In SI system, the unit of time is s and the unit of distance is m
• So the units of k can be calculated as: $\mathbf\small{k=\frac{s^2}{m^3}}$
4. We want time in 'terms of days' and distance in 'terms of km'
• 1 s = $\mathbf\small{\frac{1}{24 \times 60 \times 60}=\frac{1}{86400}}$ days
• 1 m = 10-3 km
5. So 1 s2 m-3 = $\mathbf\small{\frac{(\frac{1}{86400})^2}{(10^{-3})^3}}$
• So 10-13 s2 m-3 = $\mathbf\small{10^{-13}\times \frac{(\frac{1}{86400})^2}{(10^{-3})^3}}$ = 1.33 × 10-14 days2 km-3
Part (ii):
• Using Eq.8.25, we get: $\mathbf\small{1.33 \times 10^{-14}\times 3.84 \times 10^5}$ = 753.08
• Thus T = ✓(753.08) = 27.3 days
• In the next section we will see energy of satellites
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