In the previous section, we saw the force exerted by earth at three important positions. We will write a summary:
Eq.8.4:
Magnitude of the gravity on a point mass mA situated on the surface of the earth:
$\mathbf\small{|\vec{F}_{G(E,A_{surface})}|=\frac{GM_E\,m_A}{R_E^2}}$
Eq.8.5:
Magnitude of the gravity on a point mass mA situated at a depth d below the surface of the earth:
$\mathbf\small{|\vec{F}_{G(E,A_{depth})}|=G\,m_A \left( \frac{4 \pi \rho }{3}\right)r}$
OR
Eq.8.6:
$\mathbf\small{|\vec{F}_{G(E,A_{depth})}|=\frac{G\,M_E\,m_A}{R_E^3}r}$
• For that we apply Newton's second law
That is., When we divide force by mass, we get acceleration
• In vector form, the magnitude of 'acceleration due to gravity' is expressed as $\mathbf\small{|\vec{g}|}$
• We will now find $\mathbf\small{|\vec{g}|}$ at the three positions:
Position 1: On the surface of the earth
1. Eq.8.4, gives the force. We must divide this force by mass. Thus we get:
Acceleration (due to gravity) experienced by a point mass mA situated on the surface of the earth
= $\mathbf\small{|\vec{g}|= \frac{GM_E\,m_A}{R_E^2}\times \frac{1}{m_A}=\frac{GM_E}{R_E^2}}$
2. Thus we can write:
Eq.8.7: $\mathbf\small{|\vec{g}|=\frac{GM_E}{R_E^2}}$
3. $\mathbf\small{|\vec{g}|}$ can be determined independently by means of experiments. In such a situation, ME is the only unknown quantity in Eq.8.7. So we can easily calculate ME
• Thus a popular statement came into existence: Cavendish weighed the Earth
(i) A mass on the surface of the earth may be stationary. But still, gravity is acting on it
(ii) It is stationary because, the reaction from the ground cancels the force of gravity
(iii) If that ground is removed, the mass will begin to travel towards the center of the earth with an acceleration of $\mathbf\small{|\vec{g}|}$
(iv) Similarly, if a mass is dropped from a moderate height above the surface of the earth, it will fall with an acceleration of $\mathbf\small{|\vec{g}|}$
• The above steps are helpful to get the idea about 'acceleration on surface of earth'
Position 2: At a height h above the surface of the earth
1. Eq.8.3 gives the gravity at any point above the surface of the earth
• If the point mass is at a height h, it's distance r from the center of the earth O will be given by:
r = (RE+h)
2. We have Eq.8.3: $\mathbf\small{|\vec{F}_{G(E,A)}|=\frac{GM_E\,m_A}{r^2}}$
3. Now we divide this force by mass. We will get the acceleration experienced by that body at height h
• We will denote this acceleration as $\mathbf\small{|\vec{g}_h|}$
• So we can write: $\mathbf\small{|\vec{g}_h|=\frac{GM_E\,m_A}{(R_E+h)^2}\times \frac{1}{m_A}}$
4. Thus we get:
Eq.8.8: $\mathbf\small{|\vec{g}_h|=\frac{GM_E}{(R_E+h)^2}}$
5. Compare Eqs.8.7 and 8.8
• The numerators are the same
• But in Eq.8.8, the denominator is larger. So we can write:
At a height h, the acceleration due to gravity is lesser than that on the surface of the earth
6. Now let us write a comparison between h and RE
(i) The radius (RE) of the earth is 6378.1 km
(ii) The fig. below shows the various layers of the atmosphere. The fig. is obtained from wikimedia commons. The link is given below:
https://commons.wikimedia.org/wiki/File:Atmosphere_layers.jpg
(iii) We see that the maximum height of the atmosphere is 90 km
• This is far less than the radius RE
(iv) So in normal cases, we can write: $\mathbf\small{h<<R_E}$
• In such cases, the ratio $\mathbf\small{\frac{h}{R_E}}$ will be much less than 1
♦ An example:
♦ Let us consider a normal case. The case of a weather balloon at 40 km above the earth's surface
♦ The ratio $\mathbf\small{\frac{h}{R_E}=\frac{40}{6378.1}=0.0063}$
• So we can write: For normal cases, $\mathbf\small{\frac{h}{R_E}<<1}$
7. Consider the denominator in Eq.8.8: $\mathbf\small{(R_E+h)^2}$
• This can be written as: $\mathbf\small{\left[R_E \left(1+\frac{h}{R_E}\right)\right]^2}$
= $\mathbf\small{R_E^2 \left(1+\frac{h}{R_E}\right)^2}$
• So Eq.8.8 becomes: $\mathbf\small{|\vec{g}_h|=\frac{GM_E}{R_E^2 \left(1+\frac{h}{R_E}\right)^2}=\left[\frac{GM_E}{R_E^2}\times \frac{1}{\left(1+\frac{h}{R_E}\right)^2} \right]}$
8. But from Eq.8.7, $\mathbf\small{\frac{GM_E}{R_E^2}}$ is $\mathbf\small{|\vec{g}|}$ on the surface of the earth
• So the result in (7) becomes:
Eq.8.9: $\mathbf\small{|\vec{g}_h|=\frac{|\vec{g}|}{\left(1+\frac{h}{R_E}\right)^2}}$
$\mathbf\small{\Rightarrow |\vec{g}_h|=|\vec{g}|\left(1+\frac{h}{R_E}\right)^{-2}}$
9. The exponent is '-2'. So we have to use binomial expansion
• We saw that, $\mathbf\small{\frac{h}{R_E}}$ will be small. So $\mathbf\small{\frac{h^2}{R_E^2}}$ will be very small
• For example, 0.001 is a small quantity. It's square (0.0012 = 0.000001) is a very small quantity
• So we can discard $\mathbf\small{\frac{h^2}{R_E^2}}$ and higher powers
10. So the result in (8) becomes:
Eq.8.10: $\mathbf\small{|\vec{g}_h|=|\vec{g}|\left(1-\frac{2h}{R_E}\right)}$
11. For small values of h, the value of $\mathbf\small{\frac{2h}{R_E}}$ will be less than 1
• So the value of $\mathbf\small{\left(1-\frac{2h}{R_E}\right)}$ will also be less than 1 but greater than zero
• So the value of $\mathbf\small{|\vec{g}|\left(1-\frac{2h}{R_E}\right)}$ will be less than $\mathbf\small{|\vec{g}|}$
• That means: the value of $\mathbf\small{|\vec{g}_h|}$ will be less than the value of $\mathbf\small{|\vec{g}|}$
■ We can also write:
For small heights h, the acceleration due to gravity decreases by a factor $\mathbf\small{\left(1-\frac{2h}{R_E}\right)}$
Position 3: At a depth d below the surface of the earth
1. Eq.8.6 gives the gravity at any point below the surface of the earth
• If the point mass is at a depth d, it's distance r from the center of the earth O will be given by:
r = (RE-d)
2. We have Eq.8.6: $\mathbf\small{|\vec{F}_{G(E,A_{depth})}|=\frac{G\,M_E\,m_A}{R_E^3}r}$
3. Now we divide this force by mass. We will get the acceleration experienced by that body at depth d
• We will denote this acceleration as $\mathbf\small{|\vec{g}_d|}$
• So we can write: $\mathbf\small{|\vec{g}_d|=\frac{G\,M_E\,m_A}{R_E^3}(R_E-d)\times \frac{1}{m_A}}$
$\mathbf\small{\Rightarrow |\vec{g}_d|=\frac{G\,M_E}{R_E^3}(R_E-d)}$
4. The above result in (3) can be written as: $\mathbf\small{|\vec{g}_d|=\frac{G\,M_E}{R_E^2}\times\frac{(R_E-d)}{R_E}}$
• But from Eq.8.7, $\mathbf\small{\frac{GM_E}{R_E^2}}$ is $\mathbf\small{|\vec{g}|}$ on the surface of the earth
• So the result in (3) becomes: $\mathbf\small{|\vec{g}_d|=|\vec{g}|\frac{(R_E-d)}{R_E}=|\vec{g}|\left(1- \frac{d}{R_E}\right)}$
5. So we get:
Eq.8.11: $\mathbf\small{|\vec{g}_d|=|\vec{g}|\left(1- \frac{d}{R_E}\right)}$
6. d will be always less than RE. So the value of $\mathbf\small{\frac{d}{R_E}}$ will be less than 1
• So the value of $\mathbf\small{\left(1-\frac{d}{R_E}\right)}$ will also be less than 1 but greater than zero
• So the value of $\mathbf\small{|\vec{g}|\left(1-\frac{d}{R_E}\right)}$ will be less than $\mathbf\small{|\vec{g}|}$
• That means: the value of $\mathbf\small{|\vec{g}_d|}$ will be less than the value of $\mathbf\small{|\vec{g}|}$
■ We can also write:
At a depth d, the acceleration due to gravity decreases by a factor $\mathbf\small{\left(1-\frac{d}{R_E}\right)}$
(i) Result in position 2 (11)
(ii) Result in Position 3 (6)
• We see that, in both cases, there will be a decrease in the acceleration. That is:
♦ $\mathbf\small{|\vec{g}_d|}$ is less than the $\mathbf\small{|\vec{g}|}$ at the surface
♦ $\mathbf\small{|\vec{g}_h|}$ is also less than the $\mathbf\small{|\vec{g}|}$ at the surface
■ We can write:
• The 'acceleration due to gravity of the earth' is maximum on the surface of the earth
• The acceleration decreases when we go higher up above the surface
• The acceleration decreases when we go lower down below the surface
Solved example 8.14
The mass of a planet P is 4 times the mass of the earth. It's radius is also 4 times the radius of the earth. What is the relation between the following two items:
(i) $\mathbf\small{|\vec{g}|}$ (ii) Acceleration (due to gravitational force) experienced by a point mass on the surface of P
Solution:
1. $\mathbf\small{|\vec{g}|}$ is simply the acceleration (due to gravitational force) experienced by a point mass on the surface of the earth
• The formula is: $\mathbf\small{|\vec{g}|=\frac{GM_E}{R_E^2}}$
2. We derived this formula by assuming that the earth is made up of concentric shells
• This assumption can be applied to any planet having a spherical shape
• So for the given planet, we can write: $\mathbf\small{|\vec{g}_P|=\frac{GM_P}{R_P^2}}$
3. Taking ratios, we get: $\mathbf\small{\frac{|\vec{g}|}{|\vec{g}_P|}=\frac{GM_E}{R_E^2}\times\frac{R_P^2}{GM_P}}$
$\mathbf\small{\Rightarrow \frac{|\vec{g}|}{|\vec{g}_P|}=\frac{M_E}{M_P}\times\frac{R_P^2}{R_E^2}}$
$\mathbf\small{\Rightarrow \frac{|\vec{g}|}{|\vec{g}_P|}=\frac{M_E}{4M_E}\times\frac{(4R_E)^2}{R_E^2}}$
$\mathbf\small{\Rightarrow \frac{|\vec{g}|}{|\vec{g}_P|}=4}$
$\mathbf\small{\Rightarrow |\vec{g}_P|=\frac{|\vec{g}|}{4}}$
4. So we can write:
Acceleration (due to gravitational force) experienced by a point mass on the surface of P is one fourth of $\mathbf\small{|\vec{g}|}$
Eq.8.3:
Magnitude of the gravity on a point mass mA at any distance r:
$\mathbf\small{|\vec{F}_{G(E,A)}|=\frac{GM_E\,m_A}{r^2}}$
Magnitude of the gravity on a point mass mA at any distance r:
$\mathbf\small{|\vec{F}_{G(E,A)}|=\frac{GM_E\,m_A}{r^2}}$
Magnitude of the gravity on a point mass mA situated on the surface of the earth:
$\mathbf\small{|\vec{F}_{G(E,A_{surface})}|=\frac{GM_E\,m_A}{R_E^2}}$
Eq.8.5:
Magnitude of the gravity on a point mass mA situated at a depth d below the surface of the earth:
$\mathbf\small{|\vec{F}_{G(E,A_{depth})}|=G\,m_A \left( \frac{4 \pi \rho }{3}\right)r}$
OR
Eq.8.6:
$\mathbf\small{|\vec{F}_{G(E,A_{depth})}|=\frac{G\,M_E\,m_A}{R_E^3}r}$
• Our next aim is to find the acceleration experienced by the point mass at the three positions
That is., When we divide force by mass, we get acceleration
• In vector form, the magnitude of 'acceleration due to gravity' is expressed as $\mathbf\small{|\vec{g}|}$
• We will now find $\mathbf\small{|\vec{g}|}$ at the three positions:
Position 1: On the surface of the earth
1. Eq.8.4, gives the force. We must divide this force by mass. Thus we get:
Acceleration (due to gravity) experienced by a point mass mA situated on the surface of the earth
= $\mathbf\small{|\vec{g}|= \frac{GM_E\,m_A}{R_E^2}\times \frac{1}{m_A}=\frac{GM_E}{R_E^2}}$
2. Thus we can write:
Eq.8.7: $\mathbf\small{|\vec{g}|=\frac{GM_E}{R_E^2}}$
3. $\mathbf\small{|\vec{g}|}$ can be determined independently by means of experiments. In such a situation, ME is the only unknown quantity in Eq.8.7. So we can easily calculate ME
• Thus a popular statement came into existence: Cavendish weighed the Earth
■ We wrote: Acceleration (due to gravity) experienced by a point mass mA situated on the surface of the earth
(ii) It is stationary because, the reaction from the ground cancels the force of gravity
(iii) If that ground is removed, the mass will begin to travel towards the center of the earth with an acceleration of $\mathbf\small{|\vec{g}|}$
(iv) Similarly, if a mass is dropped from a moderate height above the surface of the earth, it will fall with an acceleration of $\mathbf\small{|\vec{g}|}$
• The above steps are helpful to get the idea about 'acceleration on surface of earth'
1. Eq.8.3 gives the gravity at any point above the surface of the earth
• If the point mass is at a height h, it's distance r from the center of the earth O will be given by:
r = (RE+h)
2. We have Eq.8.3: $\mathbf\small{|\vec{F}_{G(E,A)}|=\frac{GM_E\,m_A}{r^2}}$
3. Now we divide this force by mass. We will get the acceleration experienced by that body at height h
• We will denote this acceleration as $\mathbf\small{|\vec{g}_h|}$
• So we can write: $\mathbf\small{|\vec{g}_h|=\frac{GM_E\,m_A}{(R_E+h)^2}\times \frac{1}{m_A}}$
4. Thus we get:
Eq.8.8: $\mathbf\small{|\vec{g}_h|=\frac{GM_E}{(R_E+h)^2}}$
5. Compare Eqs.8.7 and 8.8
• The numerators are the same
• But in Eq.8.8, the denominator is larger. So we can write:
At a height h, the acceleration due to gravity is lesser than that on the surface of the earth
6. Now let us write a comparison between h and RE
(i) The radius (RE) of the earth is 6378.1 km
(ii) The fig. below shows the various layers of the atmosphere. The fig. is obtained from wikimedia commons. The link is given below:
https://commons.wikimedia.org/wiki/File:Atmosphere_layers.jpg
(iii) We see that the maximum height of the atmosphere is 90 km
• This is far less than the radius RE
(iv) So in normal cases, we can write: $\mathbf\small{h<<R_E}$
• In such cases, the ratio $\mathbf\small{\frac{h}{R_E}}$ will be much less than 1
♦ An example:
♦ Let us consider a normal case. The case of a weather balloon at 40 km above the earth's surface
♦ The ratio $\mathbf\small{\frac{h}{R_E}=\frac{40}{6378.1}=0.0063}$
• So we can write: For normal cases, $\mathbf\small{\frac{h}{R_E}<<1}$
7. Consider the denominator in Eq.8.8: $\mathbf\small{(R_E+h)^2}$
• This can be written as: $\mathbf\small{\left[R_E \left(1+\frac{h}{R_E}\right)\right]^2}$
= $\mathbf\small{R_E^2 \left(1+\frac{h}{R_E}\right)^2}$
• So Eq.8.8 becomes: $\mathbf\small{|\vec{g}_h|=\frac{GM_E}{R_E^2 \left(1+\frac{h}{R_E}\right)^2}=\left[\frac{GM_E}{R_E^2}\times \frac{1}{\left(1+\frac{h}{R_E}\right)^2} \right]}$
8. But from Eq.8.7, $\mathbf\small{\frac{GM_E}{R_E^2}}$ is $\mathbf\small{|\vec{g}|}$ on the surface of the earth
• So the result in (7) becomes:
Eq.8.9: $\mathbf\small{|\vec{g}_h|=\frac{|\vec{g}|}{\left(1+\frac{h}{R_E}\right)^2}}$
$\mathbf\small{\Rightarrow |\vec{g}_h|=|\vec{g}|\left(1+\frac{h}{R_E}\right)^{-2}}$
9. The exponent is '-2'. So we have to use binomial expansion
• We saw that, $\mathbf\small{\frac{h}{R_E}}$ will be small. So $\mathbf\small{\frac{h^2}{R_E^2}}$ will be very small
• For example, 0.001 is a small quantity. It's square (0.0012 = 0.000001) is a very small quantity
• So we can discard $\mathbf\small{\frac{h^2}{R_E^2}}$ and higher powers
10. So the result in (8) becomes:
Eq.8.10: $\mathbf\small{|\vec{g}_h|=|\vec{g}|\left(1-\frac{2h}{R_E}\right)}$
11. For small values of h, the value of $\mathbf\small{\frac{2h}{R_E}}$ will be less than 1
• So the value of $\mathbf\small{\left(1-\frac{2h}{R_E}\right)}$ will also be less than 1 but greater than zero
• So the value of $\mathbf\small{|\vec{g}|\left(1-\frac{2h}{R_E}\right)}$ will be less than $\mathbf\small{|\vec{g}|}$
• That means: the value of $\mathbf\small{|\vec{g}_h|}$ will be less than the value of $\mathbf\small{|\vec{g}|}$
■ We can also write:
For small heights h, the acceleration due to gravity decreases by a factor $\mathbf\small{\left(1-\frac{2h}{R_E}\right)}$
Position 3: At a depth d below the surface of the earth
1. Eq.8.6 gives the gravity at any point below the surface of the earth
• If the point mass is at a depth d, it's distance r from the center of the earth O will be given by:
r = (RE-d)
2. We have Eq.8.6: $\mathbf\small{|\vec{F}_{G(E,A_{depth})}|=\frac{G\,M_E\,m_A}{R_E^3}r}$
3. Now we divide this force by mass. We will get the acceleration experienced by that body at depth d
• We will denote this acceleration as $\mathbf\small{|\vec{g}_d|}$
• So we can write: $\mathbf\small{|\vec{g}_d|=\frac{G\,M_E\,m_A}{R_E^3}(R_E-d)\times \frac{1}{m_A}}$
$\mathbf\small{\Rightarrow |\vec{g}_d|=\frac{G\,M_E}{R_E^3}(R_E-d)}$
4. The above result in (3) can be written as: $\mathbf\small{|\vec{g}_d|=\frac{G\,M_E}{R_E^2}\times\frac{(R_E-d)}{R_E}}$
• But from Eq.8.7, $\mathbf\small{\frac{GM_E}{R_E^2}}$ is $\mathbf\small{|\vec{g}|}$ on the surface of the earth
• So the result in (3) becomes: $\mathbf\small{|\vec{g}_d|=|\vec{g}|\frac{(R_E-d)}{R_E}=|\vec{g}|\left(1- \frac{d}{R_E}\right)}$
5. So we get:
Eq.8.11: $\mathbf\small{|\vec{g}_d|=|\vec{g}|\left(1- \frac{d}{R_E}\right)}$
6. d will be always less than RE. So the value of $\mathbf\small{\frac{d}{R_E}}$ will be less than 1
• So the value of $\mathbf\small{\left(1-\frac{d}{R_E}\right)}$ will also be less than 1 but greater than zero
• So the value of $\mathbf\small{|\vec{g}|\left(1-\frac{d}{R_E}\right)}$ will be less than $\mathbf\small{|\vec{g}|}$
• That means: the value of $\mathbf\small{|\vec{g}_d|}$ will be less than the value of $\mathbf\small{|\vec{g}|}$
■ We can also write:
At a depth d, the acceleration due to gravity decreases by a factor $\mathbf\small{\left(1-\frac{d}{R_E}\right)}$
■ Let us compare the two results:
(ii) Result in Position 3 (6)
• We see that, in both cases, there will be a decrease in the acceleration. That is:
♦ $\mathbf\small{|\vec{g}_d|}$ is less than the $\mathbf\small{|\vec{g}|}$ at the surface
♦ $\mathbf\small{|\vec{g}_h|}$ is also less than the $\mathbf\small{|\vec{g}|}$ at the surface
■ We can write:
• The 'acceleration due to gravity of the earth' is maximum on the surface of the earth
• The acceleration decreases when we go higher up above the surface
• The acceleration decreases when we go lower down below the surface
Now we will see a solved example
The mass of a planet P is 4 times the mass of the earth. It's radius is also 4 times the radius of the earth. What is the relation between the following two items:
(i) $\mathbf\small{|\vec{g}|}$ (ii) Acceleration (due to gravitational force) experienced by a point mass on the surface of P
Solution:
1. $\mathbf\small{|\vec{g}|}$ is simply the acceleration (due to gravitational force) experienced by a point mass on the surface of the earth
• The formula is: $\mathbf\small{|\vec{g}|=\frac{GM_E}{R_E^2}}$
2. We derived this formula by assuming that the earth is made up of concentric shells
• This assumption can be applied to any planet having a spherical shape
• So for the given planet, we can write: $\mathbf\small{|\vec{g}_P|=\frac{GM_P}{R_P^2}}$
3. Taking ratios, we get: $\mathbf\small{\frac{|\vec{g}|}{|\vec{g}_P|}=\frac{GM_E}{R_E^2}\times\frac{R_P^2}{GM_P}}$
$\mathbf\small{\Rightarrow \frac{|\vec{g}|}{|\vec{g}_P|}=\frac{M_E}{M_P}\times\frac{R_P^2}{R_E^2}}$
$\mathbf\small{\Rightarrow \frac{|\vec{g}|}{|\vec{g}_P|}=\frac{M_E}{4M_E}\times\frac{(4R_E)^2}{R_E^2}}$
$\mathbf\small{\Rightarrow \frac{|\vec{g}|}{|\vec{g}_P|}=4}$
$\mathbf\small{\Rightarrow |\vec{g}_P|=\frac{|\vec{g}|}{4}}$
4. So we can write:
Acceleration (due to gravitational force) experienced by a point mass on the surface of P is one fourth of $\mathbf\small{|\vec{g}|}$
• We have seen the basics about gravitational force, gravity and acceleration due to gravity
• In the next section, we will see a few more solved examples
• In the next section, we will see a few more solved examples
No comments:
Post a Comment