# Force

1

a. Acceleration due to gravity:

Ans: The acceleration produced in a a freely falling body due to the influence of gravity is called Acceleration due to gravity.

Its symbol is “g” and its SI unit is meter per second (m/s2).

b. Weightlessness:

Ans: Weightlessness is the condition at which the apparent weight of a body is zero. Weightlessness is possible in the absence of gravity.

2

a

Ans:

 Gravity Gravitation It is the force which pulls an object towards the centre of the earth. It is the force with which two objects attract each other. It depends upon mass and radius of planet. It depends on the mass of two objects and distance between their centers.

b.

Ans:

 Mass Weight It is the quantity of matter contained in the body. It is the force with which the object is attracted towards the centre of the earth. It is scalar quantity It is vector quantity.

3.

Ans: Newton’s Law of Gravitation 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 centers.”

4.

a.

Ans:

When a coin and a feather are dropped from the roof of the building, the coin falls on the ground earlier because as the weight of coin is larger than that of the feather as well as the air resistance.

b.

Ans:

Newton’s law of the gravitation holds true or applicable for all the objects present in this universe whether the objects be terrestrial or celestial. The gravitational force exists everywhere in this universe. Therefore Newton’s law of gravitation is called universe law.

c.

Ans:

We have relation:

Weight=mass * gravity

As gravity decreases when the object is far from the surface of the earth so weight of the object also decrease when distance is increased.

d.

Ans:

Weight depends upon gravity and gravity depends on the distance.

Weight=mass * gravity

As the stone is away from the earth surface the weight of the stone is less than the stone present in the bottom.

e.

Ans:

We have relation:

Weight=mass * gravity

The value of gravity on the moon is six times less than that of the earth so the weight of object is less in moon than in earth.

f.

Ans:

As there is less gravitational force on the moon it does not attract the atmosphere on its surface as a result there is no atmosphere on the surface of the moon.

g.

Ans:

We have relation:

Weight = mass * gravity

When the body reaches the under the earth gravity decreases as a result the body will have less mass.

h.

Ans:

The value of g is greater at the poles than that of the equator because gravity depends on radius of the earth and as the equator is nearer to the centre of earth the value is more at the equator than the poles.

i.

Ans:

The person does not get injured when he falls from the parachute on the earth from the plane because there is presence of air resistance and the body does not fall freely.

5.

Ans:

The relation of gravitational force between two heavenly bodies with the product of their masses is directional proportional and inversely proportional to the distance between their centers.

6

Ans:

Newton’s gravitational constant (G) is the gravitational existence between two bodies of unit masses placed at a unit distance from their centers.

The numerical value of gravitational constant (G) is 6.67*10-11 and its unit is Nm2/Kg2.

7.

Ans: A coin and feather when dropped in vacuum falls simultaneously due to free fall as there is no external resistance in vacuum.

8.

Ans:

The body will have more weight on the equator because gravity depends on distance as the equator is closer to the centre of earth than the poles so the body will have more weight on the equator.

9.

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The attraction of 5Kg of iron to the earth is greater because the earth has more mass than the iron and the earth tends to pull the iron to its surface.

10.

Ans:

The statement is true because there occurs gravitational pull between apple and the earth. As the mass of apple is negligible as compared to the earth, the earth movement is not noticeable.

11.

Ans:

The value of G is 6.67*10-11 Nm2/kg2.

 Gravity Gravitation It is the force which pulls an object towards the centre of the earth. It is the force with which two objects attract each other. It depends upon mass and radius of planet. It depends on the mass of two objects and distance between their centers.

12.

Ans:

On the basis of his experiments , Galileo concluded that the acceleration of a an object falling freely towards the earth does not depend on the mass of the   object .To verify this fact Robert Boyle performed an experiment in a vacuum. He put a feather and a coin in a tall jar and air from the jar was removed by using vacuum pump. When the jar was inverted both the coin and feather fell to the bottom simultaneously in the absence of medium.

13.

Ans:

According to the Newton’s law of Gravitation,

F =$\: \frac{{GMm}}{{{d^2}}}$

=$\: \frac{{GMm}}{{{{\left( {\frac{d}{2}} \right)}^2}}}$

=$\: \frac{{4*GMm}}{{{d^2}}}$

Or, F = 4 * ($\frac{{GMm}}{{{d^2}}}$)

Hence, when the distance between the centers of the two body is halved the gravitational force increases by 4 times.

14.

Ans:

According to the Newton’s law of Gravitation,

F =$\: \frac{{G*2M*2m}}{{{d^2}}}$

=$\: \frac{{4*GMm}}{{{{\left( {\frac{d}{2}} \right)}^2}}}$

=$\: \frac{{16*GMm}}{{{d^2}}}$

Or, F = 16 * ($\frac{{GMm}}{{{d^2}}}$)

Hence, when the distance between the centers of the two body is doubled and their centers is halved the gravitational force increases by 16 times.

15.

Ans:

When an astronaut is in satellite they both are in freefall towards the earth. The astronaut does not exert any force on the side of the satellite. As a result the astronaut feels to be floating weightlessly though the force of gravity at the distance may not be zero. As a result the earth attracts the astronaut but negligibly.

16.

Ans:

The importance of Gravitational force is:

a. Existence of atmosphere on the surface of the earth.

b. Flowing of river.

17.

Ans:

When an object is falling towards the surface of the earth only under the influence of gravity without external resistance, the fall of the object is free fall.

The weight of the body increases when it falls under the influence of gravity as it comes closer to the core of earth.

Numerical Problems:

1.

Solution:

Given

Mass of Sphere 1 (M) = 40 Kg

Mass of Sphere 2 (m) = 15 Kg

Force(F) = $9.8{\rm{*}}{10^{ - 7}}{\rm{N}}$

Distance between Sphere (d)= 20 cm=20/100 m = 0.2m

UniversalGravitational Constant(G) =?

We Know

Or,${\rm{F}} = \frac{{{\rm{GMm}}}}{{{{\rm{r}}^2}}}$

Or,   9.8* 10-7 =${\rm{\: }}\frac{{{\rm{G*}}40{\rm{*}}15}}{{{{0.2}^2}}}{\rm{\: }}$

Hence, G is6.5 * 10-11N/m2

2.

Solution:

Given

Mass of Earth (M) = M

Mass of heavenlybody (m) = M/2

Radius of Earth (R) = R

Radius of Heavenly Body (r) = R/2

Weight of stone on Earth (W) = 100N

Acceleration due to Gravity (g) =10 m/s2

We Know

First Part

On Earth

Weight = Mass *Gravity

Or, 100 = M*10

Or, M=10 Kg

Also,

g=$\frac{{{\rm{GM}}}}{{{{\rm{R}}^2}}}$

Or, 10 =$\frac{{\left( {6.67{\rm{\: *\: }}10 - 7{\rm{\: *\: }}10} \right)}}{{{\rm{R}}2}}$

Or R=8.167 * 10-11m

Second Part

On heavenly body

Weight = Mass* Gravity

= mg

= $\frac{{10}}{2}{\rm{*}}\frac{{\left( {6.67{\rm{*}}{{10}^{ - 11}}{\rm{*}}10} \right)}}{{{{\left( {8.167{\rm{*}}{{10}^{ - 6}}} \right)}^2}}}$

= 200N

Hence, the stone weighs 200N in heavenly body.

3.

Solution

Given

Mass of both bodyare same so,

Mass (M1) = 100 Kg

Mass (M2) = 100 Kg

Distance Between them (R) =1m

Universal GravitationalConstant (G) =?

Force of attraction (F) =?

First Part

We Know,

F=${\rm{\: }}\frac{{{\rm{GMm}}}}{{{{\rm{r}}^2}}}$

Or, F =$\frac{{6.67{\rm{\: *\: }}{{10}^{ - 7}}{\rm{\: *\: }}100{\rm{\: *\: }}100}}{{{1^2}}}$

Hence, F= 6.67*10-3N

Second Part

No the force of attraction will not be different because G, M and d are all same in both case.

4.

Solution

Given

Acceleration due to Gravity on moon (g) =1.6m/s2

Mass of Body (M) =?

We know

Gravity = $\frac{{{\rm{GM}}}}{{{{\rm{R}}^2}}}$

Hence, Mass of Moon is6.93* 1022Kg

5.

Solution

Given

Diameter of first lead sphere (D) = 20 cm

Radius (R1) = $\frac{{\rm{D}}}{2}$ cm = 10 cm = 0.1m

Diameter of second lead sphere (d) = 2 cm

Radius (R2) =  $\frac{{\rm{d}}}{2}$ cm =1 cm = 0.01m

Distance between them (r) = 100 cm = 1 m

Force of attraction (F) =?

We Know

F=$\frac{{{\rm{GMm}}}}{{{{\rm{r}}^2}}}$

=$\frac{{6.67{\rm{\: *}}{{10}^{ - 11}}{\rm{*}}11500{\rm{*}}\frac{4}{3}{\rm{*}}\frac{{22}}{7}{\rm{*}}{{0.1}^3}{\rm{*}}11500{\rm{\: *}}\frac{4}{3}{\rm{*}}\frac{{22}}{7}{\rm{*}}{{0.01}^3}}}{{{1^2}}}$(Density=$\frac{{{\rm{Mass}}}}{{{\rm{Volume}}}}$)and (Volume =$\frac{4}{3}{\rm{*pi*}}{{\rm{r}}^3}){\rm{\: \: }}$

= 1.55 * 10-10 N

Hence, force of attraction is 1.55 * 10-10 N.

6.

Solution

Given

Mass of both boys are same so,

Mass (M1) = 60 Kg

Mass (M2) = 60 Kg

Distance between them (r) =1m

Universal GravitationalConstant (G) =6.67* 10-11N/m2

Force (F) =?

We Know,

F=$\frac{{{\rm{GMm}}}}{{{{\rm{r}}^2}}}$

Or F =$\frac{{\left( {6.67{\rm{\: *\: }}{{10}^{ - 11}}{\rm{*\: }}60{\rm{\: *}}60} \right)}}{{{1^2}}}$

Hence, F = 2.4*10-7N

7.

Solution

Given

Acceleration due to Gravity (g) =9.8m/s2${\rm{\: }}$

Radius of Earth(R) =6400Km = 6400*1000m = 6400000m

Mass of Earth (M) =?

We know

g=$\frac{{{\rm{GM}}}}{{{{\rm{R}}^2}}}$

Or, 9.8=$\frac{{\left( {6.67{\rm{\: *\: }}{{10}^{ - 11}}{\rm{*M}}} \right)}}{{{{6400000}^2}}}$

Hence, M= 6.018*1024 Kg

8.

Solution

Given

Mass of Sun (M)=2* 1030 Kg

Mass of Earth (m) = 6* 1024 kg

Distance between them (r) = 1.5* 1011Kg

Universal Gravitational Constant (G) = 6.67* 10-11N/m2

Force (F) =?

We Know

F=$\frac{{{\rm{GMm}}}}{{{{\rm{r}}^2}}}$

Or F= $\frac{{6.67{\rm{*}}{{10}^{ - 11}}{\rm{\: *\: }}2{\rm{*}}{{10}^{30}}{\rm{\: *\: }}6{\rm{*}}{{10}^{24}}{\rm{\: \: }}}}{{{{\left( {1.5{\rm{*}}{{10}^{11}}} \right)}^2}}}{\rm{\: \: \: }}$

Hence, F = 3.557*1022N

9.

Solution

Given

Acceleration due to Gravity (g) = 9.8m/s2

Height (h) = 8848m

Acceleration due to Gravity on Mt Everest Peak (g2) =?

We Know,

g2 =${\rm{\: }}{\left( {\frac{{\rm{R}}}{{{\rm{R}} + {\rm{h}}}}} \right)^2}{\rm{*g}}$

Or, g2 = ${\left( {\frac{{6.4{\rm{*}}{{10}^6}}}{{6.4{\rm{*}}{{10}^6} + {\rm{\: }}8848}}} \right)^2}{\rm{\: }}$* 9.8

Hence, g2 is 9.77m/s2

10.

Solution

Given

Weight of Body (W) = 63 Kg

Radius of Earth(R) =6400Km = 6400*1000m = 6400000m

Height (h) = 3200Km = 3200000m

We know

Weight = Mass * Gravity

Or 63 = M * 10

Or M = 6.3 Kg

Also

g = $\frac{{{\rm{GM}}}}{{{{\left( {{\rm{R}} + {\rm{h}}} \right)}^2}}}$

= $\frac{{6.67{\rm{*}}{{10}^{ - 11}}{\rm{*}}6.3}}{{{{\left( {6400000 + 3200000} \right)}^2}}}$

= 4.44 m/s2

Finally

F = M * g

= 6.3 * 4.44

= 28 N

Hence, Gravitational force in that height is 28 N

11.

Solution

Given

Mass of Object in earth (M) =200 Kg

Mass of object in moon (m) = ?

Now

In Earth

Weight (W) = M * g

= 200 * 10

= 2000N

Again

In moon

Weight (W) = M * g

Or, 2000= 200 * 10

Or, 2000 = m *1.67

Or, m = 1197.60 Kg

Hence, 1197.60 Kg can be lifted in the surface of moon.

12.

Solution

Given

Mass of Jupiter (M) =2*1027 Kg

Universal Gravitational Constant (G) =6.67*10-11N/m2

Acceleration due to gravity (g) =?

We Know,

g = $\frac{{{\rm{GM}}}}{{{{\rm{r}}^2}}}$

Or, g = $\frac{{6.67{\rm{*}}{{10}^{ - 11}}{\rm{\: *\: \: }}2{\rm{*}}{{10}^{27}}}}{{{{\left( {6.5{\rm{*}}{{10}^7}} \right)}^2}}}$

Hence, g is 31.57 m/s2

Second Part

Weight (W) =?

Mass (M) = 70 Kg

We Know,

W=mg

Or W=70*31.57

Or Weight = 2209.9

Hence, Weight of the person is 2209.9N.

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