Gravitational force/gravitation: The force of attraction between any two heavenly bodies in the universe is called gravitation.
Its formula: F= (G.m1m2) /d2
Its unit: Newton (N)
Factors on which gravitational force depend:
1. Product of the masses
2. Distance between their centres.
Newton’s universal law of gravitation: It states that “The force of attraction between the two bodies is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centres.”
Consider two bodies “A” and “B” with with mass “M1” and “M2” respectively which are kept at distance “d” from their centre. The force between them is “F”. According to statement,
F ∝ M1.M2 …………………..eq i
F ∝ 1/d2 …………………..eq ii
Combining eq i and ii we get,
F ∝ (M1.M2)/d2
F = (G.M1.M2)/d2 where, G is proportionality constant and it is called gravitational constant.
Q) Why is Newton’s law of gravitation also called universal law of gravitation?
Newton’s law of gravitation is also called universal law of gravitation because it is applicable to all mass of the universe.
Q) What happens to gravitational force between two bodies if the distance between them is doubled?
Initially, F = (G.M1.M2)/d2
According to question, d=2.d
Then, F1 = (G.M1.M2)/4.d2
F1 = ¼.F
Hence, The force decreases by 4 times.
Q) What happens to gravitational force between two bodies if the distance between them is halved?
Initially, F = (G.M1.M2)/d2
According to question, d=d/2
Then, F1 = 4.(G.M1.M2)/.d2
F1 = 4.F
Hence, The force increases by 4 times.
Q) What happens to gravitational force between two bodies if the mass of one is tripled and distance between them is halved?
Initially, F = (G.M1.M2)/d2
According to question, M1=3.M1 and d=d/2
Then, F1 = 4. (G.3. M1.M2))/.d2
F1 = 12.F
Hence, The force increases by 12 times.
Gravitational constant (G): The magnitude of force of attraction between any two bodies having mass 1/1 kg each are separated through 1 meter distance is called gravitational constant (G).
Its formula: G = (F.d2)/m1m2
Its unit: Nm2/kg2
Its value: 6.67×10-11Nm2/kg2
Q) The value of G is 6.67×10-11 Nm2kg-2 . What does it mean?
It means that the gravitational force between 2 unit mass kept 1m apart is 6.67×10-11 N.
Q) When does gravitational force between 2 body becomes equal to gravitational constant?
Gravitational force between 2 bodies becomes equal to gravitational constant when two unit mass are kept 1m part.
Numerical
1) The mass of Rina is 45kg and that of Mira is 60kg. They’re sitting 5m apart. Find gravitational force between them.
Answer,
m1 = 45kg
m2 = 60kg
d = 5m
G = 6.67×10-11 Nm2kg-2
F = ?
We have,
F = (G.m1m2) / d2
F = (6.67 x 45 x 60 x 10-11 ) / 25
F = 7.2036 x 10-9 N
2) The mass of sun is 2 x 1030 kg and that of earth is 6 x 1024 kg and the distance between them is 1.5 X 108 km. Find gravitational force between them.
Answer,
m1 = 2 x 1030 kg
m2 = 6 x 1024 kg
d = 1.5 X 108 km = 1.5 X 108 x 1000 m = 1.5 X 1011 m
G = 6.67×10-11 Nm2kg-2
F = ?
We have,
F = (G.m1m2) / d2
F = (6.67 x 10-11 x 2 x 1030 x 6 x 1024) / (1.5 X 1011)2
F = (6.67 x 2 x 6 x 1024 -11+30) / (2.25 X 1022)
F = (8.04 x 1044-22) / 2.25
F = 3.557 x 1022 N
Gravitational force between the sun and the earth is 3.557 x 1022 N.
3) The mass of earth is 6 x 1024 kg and that of mars is 6 x 1023 kg. The gravitational force between them on 10th Bhadra, 2060 was 6.67 x 1016 N. Calculate the distance between them.
Answer,
m1 = 6 x 1024 kg
m2 = 6 x 1023 kg
G = 6.67×10-11 Nm2kg-2
F = 6.67×1016 N
d = ?
We have,
F = (G.m1m2) / d2
6.67×1016 = (6.67 x 10-11 x 6 x 1024 x 6 x 1023) / (d)2
(d)2 = (36 x 10-11+24+23-16)
(d)2 = (36 x 1020)
(d)2= (6 x 1010)2
d= 6 x 1010m
Distance between earth and mars was 6 x 1010m
4) The mass of earth is 6 x 1024 kg and its radius is 6380 km. Earth attracts a body on its surface on with a force of 20OON. Calculate the mass of body.
Answer,
m1 = 6 x 1024 kg
d = 6380 km = 6380 x 1000 m = 6380000 m
G = 6.67×10-11 Nm2kg-2
F = 2000 N
m2 = ?
We have,
F = (G.m1m2) / d2
2000 = (6.67 x 10-11 x 6 x 1024 x m2 ) / (6380000)2
m2 = 203.42 kg
Mass of the body is 203.42 kg
Gravity.
The force by which earth (planet/satellite) attracts any objects towards its centre is called gravity. Consider the mass of earth is ‘M’ and its radius is ‘R’ and mass of body on its surface is ‘m’
We have,
F = (G.m1m2) / d2
By condition,
m1 = M
m2 = m
d = R
Then, F = (G.M.m) / R2
Effects of gravity
- We can run, walk, work etc easily.
- Construction of building, bridges etc is possible.
- Water flows from higher to lower level.
- There is atmosphere around the earth.
- Anything thrown upward falls downwards.
- An acceleration is produced in a freely falling body.
Acceleration due to gravity (g): An acceleration produced on a freely falling object under the effect of gravity is called acceleration due to gravity.
Value of g:
- At poles of earth: 9.83 m/s2
- At equator of earth: 9.78 m/s2
- On the surface of earth: 9.8 m/s2
- On the surface of moon: 1.67 m/s2
- On the surface of Jupiter 25 m/s2
Q) Acceleration due to gravity on the surface of the earth is 9.8 m/s2. What does it mean?
It means that the velocity of freely falling object towards the surface of earth increases at the rate of 9.8 m/s in every second.
Q) Prove that acceleration due to gravity doesn’t depend on mass of falling object.
A long and wide glass cylinder is taken. A coin and feather are kept on it and lid is closed. It is now inverted It is Seen that the coin falls foster than feather as shown in figure.
But when all the air is taken out from it and its is inverted, it is seen that the feather and coin fall at same rate as shown in figure. This proves that acceleration due to gravity doesn’t depend on mass of falling object.
Conclusion of feather and coin experiment. The acceleration due to gravity(g) is same for all freely falling objects.
Prove g ∝ 1/R2
Let us consider the mass of earth is ‘M’ and its radius ‘R’. A body of mass ‘m’ is on the surface of earth.
According to Newton’s law of gravitation,
F = (G.M.m) / R2 – – – – – I
Now, from Newton’s second of motion
F = m.g – – – – – II
Equating equation I and II
m.g = (G.M.m) / R2
g = (G.M) / R2
For a heavenly body, M is constant
Hence, acceleration due gravity is inversely proportional to the square of radius of earth
Relation of g with height (H)
Let us consider the mass of earth is ‘M’ and its radius is ‘R’. An object of mass ‘m’ is at height ‘H’ from the surface of earth
Now, we have
g = (G.M) / R2
According to condition,
R = R+H
Then,
g = (G.M) / (R+H)2
Note: Value of g is 0 at outer space, null point and centre of earth.
Numerical
1) The mass of earth is 6 x 1024 kg and its radius is 6380 km. Calculate its acceleration due to gravity.
M= 6 x 1024 kg
R = 6380 km = 6380000 m
G = 6.67×10-11 Nm2kg-2
g= ?
We have,
g = (G.M) / R2
g = (6.67 x 10-11 x 6 x 1024) / (6380000)2
g = (6.67 x 6 x 1024 -11) / (4.07 X 1013)
g = (40.02 x 1013) / (4.07 X 1013)
g = 9.83 m/s2
Hence, acceleration due to gravity (g) is 9.83 m/s2
2) The mass of the earth ie 6 x 1024 kg and its radius 6.4×106 m. Find the acceleration of gravity at the top of Mt. Everest.
M= 6 x 1024 kg
R = 6.4×106 m
G = 6.67×10-11 Nm2kg-2
H = 8848 m
g= ?
We have,
g = (G.M) / (R+H)2
g = (6.67 x 10-11 x 6 x 1024) / (6.4×106 + 8848 )2
g = (6.67 x 6 x 1024 -11) / (6408848)2
g = (40.02 x 1013) / (4.107 X 1013)
g = 9.74 m/s2
Hence, acceleration due to gravity at top of Everest is 9.74 m/s2
3) The mass of the earth is 6 x 1024 kg and its radius is 6.4×106 m. Find the height from the surface of earth at which acceleration due to gravity is 4 m/s2
M= 6 x 1024 kg
R = 6.4×106 m
G = 6.67×10-11 Nm2kg-2
g= 4 m/s2
H = ?
We have,
g = (G.M) / (R+H)2
4 = (6.67 x 10-11 x 6 x 1024) / (6.4×106 + H )2
H=3602499.68 m
Height at which acceleration due to gravity is 4 is 3602499.68 m
4) The acceleration due to gravity of earth is 9.8 m/s2. The mass of Jupiter is 317 times that the mass of Earth and its radius is 11 times the radius of the earth. Calculate the acceleration due to gravity of Jupiter.
For earth,
g = (G.M) / R2
9.8 = (G.M) / R2 – – – – – I
According to question, for Jupiter
M = 317 M
R = 11 R
Then,
gj = (G x 317 M) / (11 R)2
gj = 317.G.M / 121.R2
gj = 2.61 (G.M) / R2
Replacing value from equation I we get,
gj = 2.61 x 9.8
gj = 25.57 m/s2
Hence, acceleration due to gravity of Jupiter is 25.57 m/s2
Weight (W): The Force by which earth attracts any object towards its centre is called weight.
According to Newton’s second law of motion,
W = m.g (Where, m=mass and g=acceleration due to gravity)
1) The mass of Jupiter is 1.9 x 1027 kg and its radius is 71 x 106 m. Find acceleration due to gravity of Jupiter. Also, find the weight of a body of mass 80 kg on its surface
M = 1.9 x 1027 kg
R = 71 x 106 m
G = 6.67×10-11 Nm2kg-2
m = 80 kg
g = ?
W = ?
We have,
g = (G.M) / (R)2
g = (6.67 x 10-11 x 1.9 x 1027) / (71×106 )2
g = (12.673 x 10-11+27) / (5.041×1015 )
g = (12.673 x 1016-15) / (5.041)
g = 25.139 m/s2
Again, W = m.g = 80 x 25.139 = 2011.2 N
Acceleration due to gravity on Jupiter is 25.139 m/s2 and weight of 80 kg mass is 2011.2 N
Note: Weight lifted on heavenly body = Weight lifted on earth
i.e. Wb = We
mb.gb = me.ge
Q) A weight lifter can lift 100 kg mass on earth, how much mass can he lift on moon.
me= 100kg
ge= 9.8 m/s2
gm= 1.67 m/s2
mm= ?
We have,
Weight lighted on moon = Weight lifted on earth
Wm = We
mm.gm = me.ge
mmx 1.67 = 100 x 9.8
mm = 586.83 kg
We can lift 586.83 kg on moon.
Q) Write any 4 differences between mass and weight.
| Mass | Weight |
| Mass is the total quantity of matter contained in a body. | Weight is the force by which earth attracts any object towards its centre. |
| Its unit is kilogram (kg). | Its unit is Newton (N) |
| It is measured by beam balance. | It is measured by springbalance. |
| It is constant quantity. | It varies from place to place. |
Free fall: The fall of an object due to the effect of gravity without any external resistance is called free fall.
Weightlessness: The condition in which weight of a baby appears to be 0 is called weightlessness. Conditions of weightlessness
- During free fall
- In outer space
- In null point
- At centre of the earth
Q) How does a parachutist land safely on earth?Explain.
When a parachutist jumps from a height, his velocity goes on increasing so that upward pushing force of air also increases. A condition comes in which dawnward falling force of parachute and upward pushing force of air becomes balanced and parachute falls slowly with uniform velocity. In this way, a parachutist lands safely on earth.
Solved exercise
A) Very short answer questions
a. What is geocentric theory?
The theory according to which earth lies at the centre of universe and other heavenly bodies revolve around it is called geocentric theory.
b. What is heliocentric theory?
The theory according to which sun lies at the centre and other heavenly bodies revolve around it is called heliocentric theory.
c. Who proved heliocentric theory and wha proved his claim for the first time?
Nicolaus Copernicus proposed heliocentric theory and Galileo proved his claim.
d. What are two factors that affect gravitational force?
Two factors affecting gravitational force are product of 2 masses and square of distance between centre of 2 masses.
e. Who was the scientist to measure value of ’G’ experimentally for the first time?
Henry Cavendish was the scientist to measure value of G experimentally for 1st time.
f. What is the conclusion of Galileo’s coin and feather experiment?
The conclusion of Galileo’s coin and feather experiment is that, “an acceleration produced on freely falling body doesn’t depend on its mass.”
g. What are two Factors that affect gravity?
Factors affecting gravity due to a planet are mass of body and square of radius of planet.
h. Express the relation between mass of object, heavenly body and radius of heavenly body from which the weight of an object can be calculated
W = (G.M.m) / (R)2
where,
G = Gravitational constant
M = Mass of heavenly body
m = mass of object
R = Radius of heavenly body
i. What is the condition in which acceleration due to gravity of a falling parachute is 0?
Acceleration due to gravity of falling parachutes is 0 when downward falling force of parachute and upward pushing force of air becomes equal.
j. What is the motion of an object like during a free fall above earth’s surface?
When a body has free fall above earth’s surface, its velocity goes on increasing at rate of 9.8 per second in every second and it strikes the ground with greater.
B) Short answer questions
a. Mention two obvious effects of gravitation on earth
Two obvious effects of gravitation on earth are Tide occur in sea or ocean and Earth revolves around the sun.
b. What is the force between two bodies of mass ‘X’ and ‘Y’ that lie at a distance ‘Z’
m1 = X
m2 = Y
d = Z
F = ?
We have,
F = (G.m1m2) / d2
F = (G.X.Y ) / Z2
c. Why does stone and a piece of paper dropped together farm a height not reach the earth’s surface together ?
There is air around the surface of the earth. When a stone and piece of paper are dropped together from a height, the resistance of air is more on piece of paper than on stone. Therefore, they don’t reach the surface together.
d. Why does a value of ‘g’ differ from place to place on earth?
We know that acceleration due to gravity is inversely proportional to square of radius of earth g ∝ 1/R2 . The earth isn’t completely spherical, so its radius varies from place to place. Therefore, value of ‘g’ differs on earth.
e. Why does an object weigh only one sixth in moon of that earth?
We know that mass of moon is very small than of earth so that acceleration due to gravity, of moon is 1/6th of that of earth. Due to lower value of acceleration due to gravity on moon, an object weighs only 1/6th of that of the earth.
f. It is difficult to lift a heavier stone than lighter one. Why?
We know that gravity is directly proportional to mass and mass of heavier stone is more than the lighter one. Therefore, it’s difficult to lift heavier stone than lighter one.
g. Why does the weight of an object is greater at.poles than equator?
We know acceleration due to gravity (g) is inversely proportional to square of radius of the earth i.e. g ∝ 1/R2 and radius of earth at poles is less than at equator. Due to this, value of ‘g’ is more at poles than equator. Therefore, object’s weight is greater at pales as w∝g.
C) Answer the questions:
a. Why does an object weigh mare in Terai than at top of Mt Everest?
We know that acceleration due to gravity is inversely proportional to square of radius of the earth i.e g ∝ 1/R2 and radius is less at Terai than at Everest. Therefore, object weighs mare in Terai as w∝g.
b. A Satellite orbiting around the earth is always at the state of free fall? Why?
When a satellite is orbiting around the earth, it is in circular motion due to centripetal force by gravity. Therefore, satellite orbiting the earth is always at state of free fall.
c. The mutual force of attraction between an apple and the earth is equal and opposite, but the apple falls down to the earth and the earth appears to be at rest. Why?
This is because the mass of apple is much smaller than mass of earth due to this, the force can produce the motion on apple not on the earth.
d. One can jump 6 times higher on moon on earth. Why?
The mass of the moon is much smaller than mass of earth that acceleration due to gravity on moon is only 1/6th of that of the earth. Because of this lower value of ‘g’, one can jump 6 times higher on moon than on earth.
e. It is dangerous to jump from a significant height. Why?
When we jump from a significant height, our velocity goes on increasing at rate of 9.8 m/s2 every second and we reach the ground with very high velocity so that we get injured. Therefore, it’s dangerous to jump from a significant height.
f. Differentiate.
| Gravitational force | Acceleration due to gravity |
| It is mutual force of attraction between 2 bodies. | It is acceleration produced on a freely falling body under effect of gravity. |
| Its unit is Newton (N). | Its unit is m/s2 |
| It is scalar quantity. | It is vector quantity. |
| g | G |
| It is acceleration produced on a freely falling body due to gravity. | It is force of attraction between two bodies of unit masses each separated by unit distance between their centres. |
| It is vector quantity. | It is scalar quantity. |
| Its unit is m/s2 | Its unit is Nm2kg-2 |
| Gravity | Gravitation |
| It is the force by which earth attracts objects towards its centre. | It is the mutual attraction between 2 bodies. |
| It is not universal force. | It is universal force. |
| It is vector quantity. | It is scalar quantity. |
g. Tides occur in oceans but not in lake.
We know that gravitational force is directly proportional to product of 2 mass. Mass of water is much greater in ocean than in lake. So, the gravitational force on ocean is much greater than in lake. Therefore, tides occur in oceans but not in lake.
h. Role of moon is greater than that of the sun in occurence of tide
We know that gravitational force is inversely proportional to square of distance between 2 objects. The distance between moon and earth is much less than between sun and earth. Hence, gravitational force between moon and water in ocean is much greater than between sun and ocean. Therefore, role of moon is greater than that of sun in occurrence of tides.
i. The earth has atmosphere but the moon does not
Mass of the earth is much greater than the mass of moon. So, the gravity of earth is also much greater than gravity of moon. Hence gravity of earth can hold atmosphere around it but not by moon. Therefore, moon doesn’t have atmosphere.
j. Weight of a body is more to the foot of mountain than at the top.
We know that, acceleration due to gravity is inversely proportional to square of the radius of earth i.e g ∝ 1/R2 . The radius of earth is less in foot of mountain that at its top. Hence, acceleration due to gravity is more at foot of mountain than at its top. Therefore weight of body is more a foot of mountain than at top as w∝g.
k. Weight of body reduces in mine
When we go deep down in the mine, value of ‘g’ decrease due to decrease in effective mass of the earth. Therefore, weight is body reduce in mine as w∝g.
l. Value of ‘g’ is mare on poles than at equator.
We know that ‘g’ is inversely proportional to square of radius of earth i.e. g ∝ 1/R2 , radius of earth is less in equator than at poles. Therefore,, value of ‘g’ is more on poles than equator.
m. It is not possible to use parachute on moon .
There is no air on moon so that parachute doesn’t open and it falls freely with very high velocity. Therefore, it is not possible to use parachute at moon.
n. Value of weight varies from place to place on earth’s surface.
We know that acceleration due to gravity (g) is inversely proportional to square of radius of the earth i.e. g ∝ 1/R2 , Since the earth is not completely spherical, its radius varies from one place to another and ‘g’ also varies. Value of weight varies from place to place on earth surface as w∝g.
o. Mass of body is constant but weight varies place to place.
Mass of body depends on no. of atoms or molecules which remain same everywhere but weight of body depends on acceleration due to gravity (g) which varies from place to place. Therefore, mass of body is constant but weight isn’t.
p. An object falling from a height at poles reaches foster on the ground then at equator.
We know that acceleration due to gravity (g) is inversely proportional to square of radius of the earth i.e. g ∝ 1/R2 , Radius of the earth is less at the poles than at the equator. So that (g) is more at the poles than at equator. Therefore, an object falling from a height at pole reaches foster on ground than at equator.
q. A coin and a feather dropped from a height reaches together in vacuum.
This is because acceleration due to gravity is same for all freely falling bodies
Solved numerical problems
1. If mass of A is 10 kg and that of B is 10 kg and the distance between them is 10m. Find gravitational force between them.
m1 = 10 kg
m2 =10 kg
d = 10 m
G = 6.67×10-11 Nm2kg-2
F = ?
We have,
F = (G.m1m2) / d2
F = (6.67 x 10-11 x 10 x 10) / (10)2
F = 6.67 x 10-10 N
2. What effect on F will be experienced if B is tripled by keeping other values constant?
m1 = 10 kg
m2 =10 kg
d = 10 m
G = 6.67×10-11 Nm2kg-2
F = ?
According to question, m2 = 3 m2 = 3×10 = 30 kg
F = (G.m1m2) / d2
F = (6.67 x 10-11 x 10 x 30) / (10)2
F = 2.001 x 10-9 N
3. What effect on F will be experienced if distance is reduced by 4 times by keeping other values constant?
m1 = 10 kg
m2 =10 kg
d = d/4
G = 6.67×10-11 Nm2kg-2
F = ?
We know,
F = (G.m1m2) / d2
F1 = (G.m1m2) / (d/4)2
F1 = 16 x (G.m1m2) / d2
F1 = 16 x F
Force increase by 16 times.
4. When is the gravitational force (F) equal to value of universal gravitational constant ?
m1 = 1 kg
m2 = 1 kg
d = 1 m
F = ?
We know,
F = (G.m1m2) / d2
F = (G.1.1) / 12
F = G
F is equal to G when there is meter distance between 2 unit masses.
5. Planet Mars approached closest at a distance of 15.3×107 km from the earth on may 30 2016. If mass of the earth and mars are 6×1023 kg and 6×1024 kg respectively, what will be gravitational force between these two planets?
m1 = 6×1023 kg
m2 = 6×1024 kg
d = 15.3×107 km = 1.53×1011 m
G = 6.67×10-11 Nm2kg-2
F = ?
We have,
F = (G.m1m2) / d2
F = (6.67 x 10-11 x 6×1023 x 6×1024 ) / (1.53×1011)2
F = (6.67 x 6 x 6 x 1023-11+24 ) / (1.53×1011)2
F = 1.02×1016 N
Gravitational force between them is 1.02×1016 N
6. What force is acted upon every kg sea water by moon if its 3×105 km away from earth?
m1 = 1 kg
m2 = 6×1023 kg
d = 3×105 km = 3×108 m
G = 6.67×10-11 Nm2kg-2
F = ?
We have,
F = (G.m1m2) / d2
F = (6.67 x 10-11 x 1 x 6×1023 ) / (3×108)2
F = (40.02 x 1012) / (3×108)2
F = 4.45×10-4 N
Hence, 4.45×10-4 N force is acted upon every kg sea water by moon
7. What will be weight of 100kg mass on earth surface if its volume is compressed into size of moon.
m = 100 kg
M = 6×1024 kg
R = 1.7×105 m
G = 6.67×10-11 Nm2kg-2
g = ?
W = ?
We have,
g = (G.M) / R2
g = (6.67 x 10-11 x 6×1024 ) / (1.7×105)2
g = (40.02 x 1013) / 2.89×1010
g = 13847.75 m/s2
Now, W = m.g = 100 x 13847.75 = 1384775 N
8. When is the acceleration due to gravity of a body with mass 100g dropped from a height of 400m towards earths surface.
M = 6×1024 kg
R = 6380 km = 6380000 m
H = 400m
G = 6.67×10-11 Nm2kg-2
g = ?
We have,
g = (G.M) / (R+H)2
g = (6.67 x 10-11 x 6×1024 ) / (6380000+400)2
g = (6.67x6x1024-11 ) / (6380400)2
g = 9.83 m/s2
Hence, acceleration due to gravity is 9.83 m/s2
9. Calculate value of acceleration due to gravity of earth at height of 6400 km from earths surface where radius of earth is 6.4×103 km and value of acceleration due to gravity on surface is 9.8 m/s2
H = 6400 km = 6400000 m
R = 6.4×103 km = 6.4×106 m
g = 9.8 m/s2
g1 = ?
We know,
g1 = [(R)2 / (R+H)2] x g
g1 = [( 6.4×106)2 / ( 6.4×106+6400000)2] x 9.8
g1 = [(4.09×1013) / ( 12800000)2] x 9.8
g1 = 2.45 m/s2
Hence, acceleration due to gravity at height is 2.45 m/s2
10. Two gold spheres of radius 20cm and 40cm each are separated by 170m from their centres. Calculate gravitation between them if density of gold is 19.3 g/cm3
For first sphere,
D = 19.3 g/cm3
R1 = 20 cm = 0.20 m
V1 = (4/3) x 3.14 x r3
V1 = (4/3) x 3.14 x (20)3
V1 = 33523.8 cm3
M1 = D.V1 = (19.3 x 33523.8)/1000 = 647.009 kg
For second sphere,
D = 19.3 g/cm3
R2 = 40 cm
V2 = (4/3) x 3.14 x r3
V2 = (4/3) x 3.14 x (40)3
V2 = 268190.47 cm3
M2 = D.V2 = (19.3 x 268190.47)/1000 = 5176.08 kg
Now,
F = (G.m1m2) / d2
F = (6.67 x 10-11 x 647.009 x 5176.08 ) / (170)2
F = (22337632.2 x 10-11) / (28900)
F = 7.73×10-9 N
Hence, force between them is 7.73×10-9 N
11. Two object are separated by 20m to develop gravitational force of 3.335×10-9 N. If mass of an object is 100kg. Calculate mass of another object.
m1 = 100 kg
d = 20 m
F = 3.335×10-9 N
m2 = ?
We have,
F = (G.m1m2) / d2
3.335×10-9 = (6.67 x 10-11 x 100 x m2 ) / (20)2
m2 = 2×102 kg
Hence, mass of another object is 2×102 kg
12. The earth is compressed to keep 1/3 rd of its radius. Calculate acceleration due to gravity of new earth.
M = 6×1024 kg
G = 6.67×10-11 Nm2kg-2
g = 9.83 m/s2
We have,
g = (G.M) / R2
By question, R = R/3
g1 = (G.M) / (R/3)2
g1 = 9 x (G.M) / (R)2
g1 = 9 x g
g1 = 9 x 9.83
g1 = 88.2 m/s2
Acceleration due to gravity is 88.2 m/s2
13. The earth is compressed to the size of moon. Calculate acceleration due to gravity of newly formed earth based on data.
Mass of earth = 6×1024 kg
Mass of moon = 7.2×1022 kg
Radius of earth = 63800 km
Radius of moon = 1.7×103 km
Given that,
M = 6×1024 kg
R = 1.7×103 km = 1.7×106 m
G = 6.67×10-11 Nm2kg-2
g = ?
We have,
g = (G.M) / R2
g = (6.67 x 10-11 x 6×1024 ) / (1.7×106)2
g = (6.67x6x1024-11 ) / 2.89×1012
g = (6.67x6x1024-11-12 ) / 2.89
g = 138.47 m/s2.
Hence, acceleration due to gravity of newly formed earth is 138.47 m/s2.
