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Gravity and Gravitation
Acceleration due to gravity:
The acceleration produced on a body under the influence of gravity alone is called acceleration due to gravity. It is denoted by g. The acceleration due to gravity at the surface of a planet is given by
g = 𝐺𝑀𝑅 2 , where M is mass and R is radius of the planet.
At the surface of the earth its value is found to be 9.8 m/s^2
Variation in acceleration due to gravity:
i. With height from the surface of the earth:
Consider the earth to be a sphere of mass M and radius R. The acceleration due to gravity at the surface of the earth is given by
g = 𝐺𝑀𝑅 2 ..............................(i)
Now the acceleration due to gravity at a height h from the surface of the earth is given by
g' = (^) (𝑅+ℎ)𝐺𝑀 2 ..........................(ii)
Dividing (ii) by (i), we get;
𝑔′ 𝑔 =^
𝑅^2 (𝑅+ℎ)^2
or, 𝑔
′ 𝑔 =^ (1 +^
ℎ 𝑅)
−
h
R
Expanding right hand side by using binomial theorem we get,
𝑔′ 𝑔 = 1^ –^
2ℎ 𝑅 + ... terms containing higher powers oh^
ℎ 𝑅.
For h < < R, ℎ𝑅 < < 1. So, the terms containing higher powers of (^) 𝑅ℎ can be neglected.
′ 𝑔 = 1^ –^
2ℎ 𝑅 or, ( h < < R )
This shows that the value of acceleration due to gravity goes on decreasing (g' < g) with increase in height from the surface of the earth.
ii. With depth from the surface of the earth:
Consider the earth to be a sphere of mass M and radius R. The density of the earth is given by
𝜌 = 4 𝑀 3 𝜋𝑅^3
...............................(i)
The acceleration due to gravity at the surface of the earth is given by
g = 𝐺𝑀𝑅 2 = (^) 𝑅𝐺 243 𝜋𝑅^3 𝜌
or, g = 43 𝜋𝜌𝐺𝑅 ..................................(ii)
Now the acceleration due to gravity at a depth d from the surface of the earth is only due to shaded portion of the earth of radius R-d, which is given by
g' = g (1 – 2ℎ 𝑅 )
d
R
If F be the gravitational force experienced by a mass m at any point in the gravitational field, the gravitational field intensity t that point is given by
E = (^) 𝑚𝐹 ( Its SI unit is N/kg )
Consider the earth to be a sphere of mass M and radius R. The gravitational force experienced by a mass m at a distance r from the centre of the earth is given by
F = 𝐺𝑀𝑚𝑟 2
∴ The gravitational field intensity at a distance r from the centre of the earth is given by
E = (^) 𝑚𝐹
or, E = 𝐺𝑀𝑟 2
At the surface of the earth, r = R;
∴ E = 𝐺𝑀𝑅 2
⇒ E = g
Thus gravitational field intensity is numerically equal to the acceleration due to gravity.
Gravitational Potential :
The gravitational potential at any point inside the gravitational field is defined as the amount of work done in bringing a unit mass from infinity to that point. It is a scalar quantity and denoted by V. If W be the amount of work done in bringing a mass m from infinity to any point inside the gravitational field, the gravitational potential at the point is given by
V = 𝑊𝑚 ( Its SI unit is J/kg )
m
M r
R
Expression for gravitational potential:
Consider the earth to be a sphere of mass M and radius R. Let P be a point at a distance r from the centre of the earth where the gravitational potential is to be determined. Let a unit mass is at a point at distance x from the centre of the earth. The gravitational force experienced by the unit mss t is given by
F = 𝐺𝑀𝑥 2
Now the small amount of work done in bringing the unit mass from point A to B through infinitely small distance dx is given by
dw = F dx
or dw = 𝐺𝑀𝑥 2 dx ...........................(i)
Now the total amount of work done in bringing the unit mass from infinity to point P is obtained by integrating equation (i) between the limits ∞ to r.
i.e. W = ∫∞ 𝑟𝐺𝑀𝑥 2 dx
or, W = GM∫∞ 𝑟 𝑥−2^ dx= GM [− (^1) 𝑥 ]∞𝑟^ = - GM[ (^1) 𝑟 - (^) ∞^1 ]
or, W = - 𝐺𝑀𝑟
This amount of work is called gravitational potential at P and denoted by V.
i.e. V = - 𝑮𝑴𝒓 ...............................(ii)
dx
B A P ∞
M
x
r
i.e. U = - 𝐆𝐌𝐦𝐫 ...............................(ii)
Escape Velocity:
Escape velocity is defined as the minimum velocity with which a body has to be projected vertically upward so as to escape the gravitating body.
Consider a body of mass m is projected vertically upward from the surface of the earth with velocity ve so that it can go beyond the gravitational field of the earth. Let at any instant during its upward motion, the projected mass is at point P at a distance x from the centre of the earth. The gravitational force between the earth and the projected mass at P is given by
F = 𝐺𝑀𝑚𝑥 2
Now the small amount of work done in taking the mass from point P to Q through infinitely small distance dx is given by
dw = F dx
or dw = 𝐺𝑀𝑚𝑥 2 dx ...........................(i)
Now the total amount of work done in taking the mass from the surface of the earth to infinity is obtained by integrating equation (i) between the limits R to ∞.
Q
x
∞
P
M R
dx
i.e. W = (^) ∫𝑅 ∞𝐺𝑀𝑚𝑥 2 dx
or, W = GM𝑚 ∫𝑅 ∞ 𝑥−2^ dx = GMm [− (^) 𝑥^1 ]𝑅∞^ = - GMm [ (^) ∞^1 - (^1) 𝑅 ]
or, W = 𝐺𝑀𝑚𝑅
This amount of work is done at the cost of initially supplied kinetic energy of the body.
i.e. 12 mve^2 = 𝐺𝑀𝑚𝑅
∴ ve = √2𝐺𝑀𝑅 ................................(ii)
Also using g = 𝐺𝑀𝑅 2 ⇒ GM = gR^2 we get
ve = (^) √2𝑔𝑅 .................................(ii)
Equations (i) and (ii) are the expressions for escape velocity.
For earth, g = 9.8 m/s^2 and R = 6.4 × 10^6 m,
∴ ve = √2 × 9.8 × 6.4 × 10^6 = 11 200 m/s = 11 .2 km/s
Satellites: The heavenly bodies which revolve round the planet are called satellites. Satellite may be natural or artificial. For example moon is the natural satellite of earth. Artificial satellites are used for different purposes like weather forecasting, in telecommunications, in transmitting radio TV signals etc.
Orbital velocity and time period of a satellite:
Orbital Velocity:
The velocity of a satellite in its orbit is called orbital velocity.
m
V
r
M
Using g = 𝐺𝑀𝑅 2 ⇒ GM = gR^2 , we get
T = 2𝜋√ 𝑟
3 𝑔𝑅^2
or T = 2𝜋𝑅 √𝑟
3 𝑔
or, T = 𝟐𝝅𝑹 √(𝑹+𝒉)
𝟑 𝒈 .......................(iv)
This is the expression for time period of a satellite.
Geostationary Satllites:
The satellite which appears stationary to an observer on the earth is called geostationary satellite. The orbit of a geostationary satellite is called parking orbit or geosynchronous orbit. For a satellite to be geostationary, it must satisfy the following conditions.
i. The orbit of the satellite must lie on the equatorial plane of the earth. ii. Its period and direction of revolution must be same as the period and direction of rotation of earth about its axis.
Note:
All communication satellites are geostationary satellites The minimum number of geostationary satellites needed to be placed in a geostationary orbit for worldwide communication between any two locations must be three. The period of revolution of geostationary satellite is 24 hours. The height of geostationary satellite is 36000 km and its orbital velocity is 3.1 km/s.
Total Energy of Satellite:
A satellite revolving round the earth has both kinetic and potential energy. So, the total energy of a satellite is given by the sum of its kinetic and potential energy.
𝜔
satellite earth
Fig : Geostationary Satellite
KE of a satellite:
Consider a satellite of mass m is revolving round the earth in a circular path of radius r with orbital velocity vo. The centripetal force for the satellite to revolve round the earth is provided by the gravitational force between the earth and the satellite.
i.e. Fc = Fg
or, 𝑚vo
2 𝑟 =^
𝐺𝑀𝑚 𝑟^2 , where M is mass of the earth
or, vo = √GMr ........................(i)
Now kinetic energy of the satellite is given by
KE = 12 mvo^2
∴ KE = GMm2r .........................(ii)
PE of a satellite:
The gravitational potential energy of a satellite is given by
U = - GMmr ......................(iii)
Now the total energy of a satellite is given by the sum of its KE and PE.
i.e. E = KE + PE = GMm2r + ( - GMmr )
or, E = - 𝐆𝐌𝐦𝟐𝐫 .................(iv)
Thus the total energy of a satellite is negative. This shows that the satellite is bound to the gravitational field of the earth. The total energy required to make the satellite free from the gravitational field of the earth is given by, E = GMm2r.
Global Positioning System (GPS) :
The Global Positioning System (GPS) is the Global Navigation Satellite System (GNSS) that provides location velocity and time synchronization. It uses satellites, a receiver and algorithms to synchronize location, velocity and time data for air, sea and land travel. GPS works on any weather conditions, anywhere in the world, 24 hours a day. The GPS system is created and maintained by the United States government. It is freely accessible to anyone with a GPS receiver.
Applications of GPS:
The main applications of GPS are: i. Location : It is used for determining Position. ii. Navigation : It is used to get navigation from one place to another. iii. tracking : It is used for tracking personal or object movement. iv. Maping : It is used for creating map of the world. v. Timing : It is used to make precise time measurement.
Conceptual Questions:
- Show that a satellite orbiting the earth is in the state of free fall. Ans: A body is said to be in the state of free fall if its acceleration is equal to the acceleration due to gravity. The centripetal force for the satellite to revolve around the earth is provided by the gravitational force between the earth and the satellite. i. e. Fc = Fg or, 𝑚vo
2 𝑟 =^
𝐺𝑀𝑚 𝑟^2 or, vo
2 𝑟 =^
𝐺𝑀 𝑟^2 ⇒ a = g Thus the acceleration (centripetal acceleration) of a satellite is equal to the acceleration due to gravity at the position of satellite and its direction is also towards the centre of the earth. This shows that a satellite orbiting the earth is in the state of free fall.
- When will you attract the sun more, today at noon or tonight at midnight? Justify your answer.
- A person sitting in an artificial satellite of the earth feels weightlessness, but a person standing on moon has weight through the moon is a satellite of the earth. Why?
- Suppose the radius of the earth were to shrink by 2%, its mass remaining the same, would the acceleration due to gravity 'g' increase or decrease and by what percent?
- An astronaut releases a spoon out of a satellite in the space. Will the spoon fall on the earth?
- How does the weight of a body vary on going from earth to moon? Would its mass change?
- Distinguish between gravitational potential and gravitational field strength.
- An astronaut inside a small space ship orbiting around the earth does not experience any gravity but an astronaut on the moon which is also orbiting around the earth does experience gravity. Why?
- An astronaut in a space capsule orbiting the earth experiences weightlessness. Why?
- At what condition does a body become weightless at the equator?
- Write the unit and dimension of gravitational field intensity with its definition.
- If heavier bodies are attracted more strongly by the earth why do they not fall faster than lighter ones? (neglect air resistance)
- Suppose the radius of the earth were to shrink by 2%, its mass remaining the same, calculate the percent change in the acceleration due to gravity 'g'
- If the force of gravity acts on all bodies in proportion to their masses, why does not a heavy body fall faster than a light body?
- What do you mean by geo-stationary satellite? Explain.
- Explain why the moon has no atmosphere?
- How does 'g' at a point vary with the distance from the centre of the earth? Where is the highest value of g? Explain.
- The weight of a body is less inside the earth than on the surface. Explain.
- At what depth from the surface of the earth the value of acceleration due to gravity is two third of its value at the surface of the earth?
- Can we put geostationary satellite directly above Dhangadhi?
- Which takes more fuel to take a rocket from earth to moon or moon to earth?
- Since the moon is continuously being attracted towards the centre of the earth by gravitational attraction, why does not it crash into the earth?
Numerical Problems:
- Calculate the amount of work done to move 1 kg mass from the surface of the earth to a point 10^5 km from the centre of the earth. Ans : 4 x 10^6 J
- An artificial satellite revolves round the earth in 3 hours in a circular orbit. Find the height of the satellite above the earth assuming earth as a sphere of radius 6370km. Ans : 4256.8 km
- A remote sensing satellite of the earth revolves in a circular orbit at a height of 250 km above the earth's surface. What is the orbital speed and period of revolution of the satellite? Ans : 7835.5 m/sec, 5314 sec
- A man can jump 1.5 m on earth. Calculate the approximate height he might be able to jump on a planet whose density is one quarter of the earth and whose radius is one third that of the earth. Ans : 18 m
- An artificial satellite revolves round the earth in 2.5 hours in a circular orbit. Find the height of the satellite above the earth assuming earth as a sphere of radius 6370 km. Ans : 3040 Km
- Obtain the value of g from the motion of the moon assuming that its period of rotation round the earth is 27 days 8 hours and the radius @ its orbit is 60.1 times the radius of the earth. Ans : 9.76 m/sec²
- The period of moon revolving under the gravitational force of the earth is 27.3 days. Find the distance of the moon from the centre of the earth if the mass of earth is 5.97x10^24 kg. Ans : 3.83 x 10^8 m
- A satellite of Kinetic energy E is revolving around the earth in a circular orbit. The minimum additional energy required to escape it from orbit into outer space is a) E b) E/√2 c) E/2 d) √2 E
- A body is thrown vertically upwards from the surface of earth of radius R with half of escape velocity. The maximum height attained by body is (Hint : Use conservation of energy & KE = 0 at height h) a) R/2 b) R/3 c) R/4 d) R/
- A body is projected with twice the escape velocity Ve from surface of earth then velocity of body at infinity will be (Hint : Use conservation of energy & PE = 0 at infinity) a) Ve b) √2 Ve c) √𝟑 Ve d) 0.5 Ve
- A geostationary satellite orbits around the earth in a circular orbit of radius 36000 km. The
time period of a satellite at 100 km from the surface of the earth will be (Hint : T ∝ 𝑟
3 (^2) ) a) 0.5 hrs b) 1 hr c) 1.8 hrs d) 4 hrs
- The depth blow the surface of the earth at which acceleration due to gravity is sme as that at a height 2 km above the surface of the earth is a) 1 km b) 2 km c) 4 km d) 8 km
- The period of revolution of earth's satellite close to the surface of the earth is 90 minutes. The time period of another earth satellite in an orbit at a distance of three earth's radii from its surface will be a) 90 min b) 90 x (^) √8 min c) 270 min d) 720 min
- If both the mass and radius of the earth decreases by 1%. a) The escape velocity would increase b) The acceleration due to gravity would increase c) The escape velocity would decrease d) The acceleration due to gravity would decrease.
- Energy required to move a body of mass m from an orbit of radius 2R to 3R is (where M = mass of earth, R = radius of earth) a) GMm12R 2 b) GMm3R 2 c) 𝐆𝐌𝐦𝟔𝐑 d) GMm8R
- A space-craft is launched in a circular orbit very close to earth. What additional velocity should be given to the space craft so that it might escape the earth's gravitational pull? (Radius of the earth = 6400 km, g = 9.8 m/s²) a) 3.25 km/s b) 11.2 km/s c) 8 km/s d) 20.2 km/s.
- If the radius of the earth suddenly decreases by 20% of its present value, the mass of the earth remaining constant, the value of the acceleration due to gravity will a) Remain unchanged b) Decrease by 36% c) Increase by 36% d) Increase by 56%
- Two spherical bodies of mass M and 5 M and radii R and 2 R respectively are released in free space with initial separation between their centers equal to 12 R. If they attract each
other due to gravitational force only, then the distance covered by the smaller body just before collision is (Hint: 𝑆 𝑆^12 = 𝑎 𝑎^12 & S 1 + S 2 = 9R) a) 7.5 R b) 4.5 R c) 3.5 R d) 2.5 R
- A body weighs 72 N on the surface of earth. What is the gravitational force on it due to earth at a height equal to half the radius of the earth from the surface? a) 72 b) 28 N c) 16 N d) 32 N
- Knowing that mass of moon is M/81, where M is the mass of earth, find the distance of the point from the moon, where gravitational field due to earth and moon cancel each other. Given that distance between earth and moon is 60 R, where R is the radius of earth a) 8 R b) 6 R c) 4 R d) 2 R
- A mass m is placed at a point B in the gravitation al field of mass M. When the mass m is brought from B to near point A, its gravitational potential energy will a) Remain unchanged b) increase c) decrease d) become zero.
- If g is the acceleration due to gravity on the earth's surface, the gain in potential energy of the body at a height equal to three times the radius R of the earth will be a) mgR b) 12 mgR c) 13 mgR d) 𝟑𝟒 mgR
- A body of mass m rises to a height h= R/5 from the surface of earth, where R is the radius of earth, If g is the acceleration due to gravity at the surface of earth, the increase in potential energy is a) (4/5) mgh b) (6/7) mgh c) (5/6) mgh d) mgh
- The gravitational potential difference between the surface of a planet and a point 100 m above its surface is 10 J kg. If the gravitational field over this range is uniform, then the work done to raise the body of 5 kg from the surface to a height 40 m is a) 10J b) 20J c) 40J d) 80J
- Two particles of equal mass go round a circle of radius R under the action of their mutual gravitational attraction. The speed of each particle is
a) V = √𝐺𝑀2𝑅 b) (^) 2𝑅^1 √ (^) 𝐺𝑀^1 c) 𝟏𝟐 √𝑮𝑴𝑹 d) √4𝐺𝑀𝑅
- The kinetic energy required to make a body move to infinity from the earth's surface is
a) Infinite b) 2 mg R c) 12 mg R d) mgR
- If the radius of the earth's orbit is made one fourth, then duration of an year will become
a) 8 times b) 4 times c) 𝟏𝟖 times d) 14 times
