**Q.1: What is the force of gravitation?**

Ans: M force due to which everybody of the ai^rse attracts every^^ other^ body is known as fcft-cet

**Q.2: Why earlier scientists could not guess about the gravitational force?**

Ans: The earlier scientists did not care about the force of gravitation. They took the phenomena happening around them as granted. They did not think logically about gja>/!tatton white5 New^

**Q.3: State and explain Newton’s law of gravitation. **

Ans.: According to Newton’siaw of gfayitation: “Everybody In One universe afeiefe every otHer body witH d forci w^tch^i^ directly proportional to the product of their masses and inversely proportional to the square of the distance between Jheir centres.”

Consider two bodies of masses mi and m2. The distance between their centres is.d.

‘Fig 5.1 Two masses attract each other witn a gravitational force of equal magnitude.

force of attraction F with which the two masses mr and m2

F oc or F oc

F oc nr»i

This is caHed Newton’s law of gravitation.

where G is the proportionality constant. It is called the universal constant of gravitation. Its value remains the same everywhere. In SI units, its value is 6.673 x 1(T11Nm2 kg’2.

**Q.4: Why we do not observe force of attraction between any two objects**

Ans: The value of universal (gravitational constant G is very small and the masses of objects around us are also very small. Therefore, the force of attraction between two bodies whose masses are In the range of tens of ‘hundred’s of kilogramme is very small and it will be very different for us to obsec/e that lorce.

**Q.5; If a body is at the surface of earth. What will be the gravitational force on the body? **

Ans: If a body is at the surface of earth, the gravitational force oh the body wffl’be ‘ equal to its weight,

Q.6: ‘ Do you attract the earth or the earth attracts you which one is attracting with a larger force? You or the earth.

Ans: According to Newton’s law of gravitation; everybody in the universe attracts every other body. So both earth and you attract each other. As the force of attraction of the earth on you is equal to your weight, you a^ atfract the ea’rth with the same force equal to your weight. Thus both you and the earth attract with the same force. ‘ .

Untt*»: Gravitation

**Q.7: Why law of gravitation is Important to us?**

Ans: All our activities are based on gravity. Without gravity, the things of the uniN^rse wi^fie iikjr© ^tfferent/ the whote of our universal system such as moon, sun, galaxies, is based on gravitation. In the absence of gravitation, the whote

**Q.8: How to law ff gravitation consistent with Newton’s third law of motion?**

Ans: According to law of gravitation, everybody in the universe attracts every other body. Keeping it in mind it is said that mass mv attracts mass m^ with a f oroe F while mass m2 also attracts mass m1 with the same force but in the opposite direction. If the force acting on rr\, is considered as action, theri the fprce acting on m2 will be the reaction. Action and reaction due to the force of gravitation are equal in magnitude but opposite in direction.

This is consistent with Newton’s third law of motion which states, to eVery action there is always an equal but opposite reaction.

**Q.9: How can you say that gravitational force is a field force?**

Ans: The weight of a body is due to the gravitational force with which the earth attracts a body. Gravitational force is a non-contact force. For example, the velocity of a thrown up, body goes on decreasing while on return, its velocity goes on increasing. This is due to the gravitational pull of the earth acting on the body whether the body is in contact with the earth or not. Such a fore* is called the field fore*. • .

**Q.10: If we go on the top of a mountain, will our weight increase or decrease?**

Ans: As g is inversely proportional to the square of the radius of earth i.e. g <x -^

Therefore, if R increases, the Value of g decreaisies and hence our weight, w = m g decreases ort mountains.

**Q.11: What is meant by gravitational field strength?**

Ans: Gravitational field exists all around the earth. This fieJ4 Wdirecjed towards the centre of the earth as shown by arrows. The , gravitational field becomes weaker and , , weaker as we go farther and farther away fromWeearth.

“In the gravitational field of the earth,

thei gravRatiqnaf *ife6f pet* lafcit massiscafled gr«^m^^^n^^m%Bm- – *52:•^SSSSK&SK^ * ^

**Q.12: How can the mass of earth be determined?**

Ans: Consider a body of mass m on the surface pt 4b^^i^^et,iTT^.pf;.th^i>^r||i {?% Me and radius of the earth be R The di,s|ance of the body from the centre of the earth wiU

According to the lavy of grayitation,

(1)

1

Fig. 5.3

But the force with which earth attracts a body towards its centre is equal to its weightvy,Therefore,

F = w = mg ……(2) ?s; ;

Gmf

or

or

or

mg

G g «-3

Me =

.(3)

Mass of the earth Me can be determined by putting the values of g, R and G

in Eq. (3), thus

Uoit-5; OravlUtlon

*6.0x

**Q.13: • Carl; you determine the mass of our moon? If yes, then what do you****need to know? ‘ **

Ans. According rwMvton’s law of gravitation

v,fi’

where. M8 is mass of earth and Re is radius of earth <

For moon “.

K* –

where Mm is mass of moon and Rm is radius of moon.

As F = w = mg,

or

l’.;or

G

where G is gravitat’onal• constant. Its value \s 6.673 x

W*.

Thus, if we know the’ value of g on the surface of moon and Hie radius of moon, then we <^*defetmin£^

**Q.14: Why the value of g varies from place to place?**

• a-

: As .g..=

where Me and G are constants, therefore

where R is distance from tfte. centre of the /earth..

WiysicstX

The acceleration due to;gravt^r.^ is inversely proportional to the square <

(R2) from the centre of tfie eartft i.e. g decreases as R increases. Due to this reason, acceleration due to. gravity g)varies from place to place.

**Q.15: Explain how the value of-g varies with aJtitijc|e?**,,t

Ans: We know that G1

9 =

where G and Me ar.e constants.

. It means the value of g is inversely proportional to the square of the radius of the earth. But it does not remain constant. It decreases with altitude.

Consider a body of mass m at an altitude h. The distance of the body from the centre of the earth becomes R + h. Therefore

9h=

(R+hf

Fig. 5,4: We*gtit of d tsody decreases as tts height” ‘ k$(gases from the sgrface of the Eatth.

.(1)’ •

According to Eq. (1), we come to know that at a1 height equal to one earth radius above the surface of the earth, g becomes one fourth of its value on the earth, < Similarly, at a distance of two earth radius above the earth’s surface, the value of g j becomes one ninth of its value on the e,arth.

**Q.16: What is a satellite?**

Ans: “An object that revolves around The moon revolves around the e^rth, so moon is a natural satellite of the earth.

**Q.17: What are artificial satellites?**

Ans: “Man made objects which revolve around the earth are called artificial

satellite.

**Q.18: What a geostationary satellites?**

Ans: “A satellite whose motion is synchronized with the rotation pf the earth is known as a geostationary satellite.”

Unlt-S: Qravltatlon

**Q.1 9: What is the usefulness of geostationary satellites?**

Ans: A geostationary satellite is very useful for worldwide; communic«tiof>!»»eather observations, navigation and other military purposes. –

**Q.20: Why communication satellitas-are stationed at gflos^tipfiary orbits?**

Ans: Communication siatellites are stationed in geostationary orbits, because in this orbit, communication satellites move around the earth with same speed as that of the earth. One communication satellite covers 120″ of longitude, so that whole of the 6sarth siiffiace can be covered by ^i^*ci3m^«ypos1tk5r^’^a1eJWes. Since these satellites seem to hover over one place on the earth, continuous communication at any place on the surface, of tjie^ earth, can be made.

Q**.21 : Find the relation for orbKal velocity with which a satellite is launched in an orbit around the earth.**

Ans:for a satellite to keep in a circular orbit is given by

This force is provided fey Jihe J^avitational force of irtbactwn befeween the earth and the satellite and is eo,ual to weight w of the satellite. Hence

Fc- •-‘ .”w ‘

or or as

mgh = vo2 = r0 =

Hlv

L ;ncj

v0 =

(1)

Equation (1 ) gives the velocity,, which ••& sa|prfii^ mt»st,j3<?§sesswh(?n launched in an orbit of radius r0 = R + h around the earth.

Ah approximation can be made for a satellite revolving close to the earth such that R»h.

**Q.22: On**

Arts: the ortiital speW<3f a satellite dip^tWs upon the earth siirface

**Q.23: Hj^v Newton’s law of gravitation helps in understanding the motion of .. satellites? **

Ans: The satellites are moving abund the eairth With ^rtan

a^oe^f^ipj^d thjstacc^le^jtio^

satellite and the earth according to Newton’s second lap of ffiifejri,,,

**MCQ FROM THE TEXTBOOK**

Encircle the correct answer from the following choices:

i.

U.

iii.

Earth’s gravitational force of attraction vanishes at

(a)6400km . ; ; “i”: H

(c) 42300 km

Value of g increases with the (a) increase in mass of the body (c) decrease in attitude

tb) infinity (d) 1000km

iv.

(b) increase in altitude % (d) none of these

The value of g at a height one earth’s radius above the surface of the earth is .’

(a) 2g (b) Y*g

(c) V9 g (d) Y*Q jf^ ‘.’; •

The value of g on moon’s surface is 1.6 ms”2. What wili be the weight of

a-10tfKg;6’6dy^ifc^

(a) 100 N (bflBON; . •

(e).1000N / s- % ;

v.

vi.

The altitude

are launched above the surface of the earth is

(a) 850km (b) 1000 Km

(c)640Qkm ‘ (d)42,300km .cHifiia,..: The orbital speed of a low orbit satellite is

(a) zero

(b) 8ms

~1

(i) c (ii).c

(iii) d

(ANSWERS /:\;^-:: ipv) b ; ‘• (v) d

satellites .

;! (vi) d

ADDITIONAL MCQ’S :

t

Encircle the correct answer from the followingclioices:

1. The first man who came up with the idea of gravity was

(a)Bohr »;

2. The idea of gravity was presented in

(a)1660

3. According to Newton’s law of universal gravitation

‘ 4. The value of universal constant of gravitation G is

(a) 6.673×10-11Nm2kg^/ (b) 6.673×10^ Nrt,? kg^

(d) 6.673×10-6^^

E^ll^^^^vr^:^

“v”. •^i (b}Nn>kg2 :\ (c)Nm2kgT2

The w

(a) centrif^l jfe

(c) gravitatJonai force

. . .

a body is due to the

(b) centrifugaJJ|5rce . ‘ (d) elastic restoring force

7. Which field is assumed to exist ail arourwlth« earth?

(a) electric field ^(byrir>a^ticlSeld” ‘ (c) gravitational field (d) electrostatic field

8. The gravitational force is also called

(a) field forcee^f^.(c) magnetic force (d) none of these

9. In the gravitational field of the earth, the gravitational force fx»r unit mass is called

(a) electric field strength , (b) magnetic field (c) gravitational field strength •’• (d) electric power

10. The Mass of earth is determined by using

v, (a) Me = ^f (b)Me = ^ ;=’ t :;!!-

•- <c)\i^^—^ -••• •;•••; . : ‘-.

11. v The mass of earth is equal to

. , ‘ (a}6x1Q19%J’;^^^^ . ; (c)6x1022kg 3d (d) SxlO2^

12. the value of g at a height Of two earth’s radius above the surf ace of the ‘. . earth is . . . ‘•

(a) 1/3 g ,. (b)1/6g

(c)i/9g lrv^ (d)1/l2g

13. The value of g at height h from the centre of the earth is given by

/; ^ Me2 (c)9h = G^|

14. The height of a geostationary satellite from the surface of the earth is

‘about-(a)40,300 km – (b)42,30^km (c)4S,300km (d)49,300km- .

15. The velocity of a geostationary satelUte with respect to the earth is

(a) 3000 km h’? , • (b) 26<K) km rr1 (c) 2200 km h’1 (d)zerb

16. A Global Positioning System consists of

(a) 6 earth satellites , <b)11earMsaltilHteS

(c) 18 earth satellites ”

Unit-5: Gravitation

23.

24.

25.

26.

27.

17. The satellites-ofGlobalPositioning System revolve arotM*d the earth

(a) once a day (b)twie0a day -‘•’• ^»*. (c) thrice a day (d) none of these ,

18. The satellites of Global Positioning System revolve around the earth :. with a sjjsedol ,

(a) 1.87 tons-1″ ,’ (b) 3.87,km s’1 (c) 5.87 km s~1 . (d) 7.87 krn s~1 .

19. Communication’ satellites complete their one rotation around the earth

(a) 12 hours f (b) 24 hours (c) 48’hours (d)96 hours

20. Earth completes its one rotation about its axis in

(a) 12 hours (b) 24 hours

(c) 48hours (d)72hours

21. The distance of moon from the Earth is nearly

(a) 38,00,00,000km (b) SiSO.OO.OOO km

(a) one day

:;(b.) 17.3’days

22.Which force keeps thfesatelHte to move around the earth?

(a) gravitational force (p) centripetal force

(c) centrifugal force (d) field force

23.A satellite revolving around very close to the earth has speed

(a)8kmsr1 – (b)16krns^1 ; I :

(c)24kms~1 ‘.’. (d)32kms-1

24.A satellite revolving around very close to the earth has speed

(a) 23000 km h”1 (b) 29000 km h~1 ‘

(c) 34000 km rf1 : (d) 37000 k-m rf1 ”

25.The average density of earth is approximately equal to

(a)550kgm-3r? (c) 1500 kgrtr3 -What is not true about g?

(a) g is different at differed place (c) g is less at poles \

(b) 1000 kg m

(b)g is greater at poles

(d) g decreases as we go higher

28. An artificial sateiWe keeps ot revolving arottnd the earth In different orbits with uniform speed due to

(a) gravitattonailpfce

29. What will be the value of G if mags of earth becomes four times?

30.

(c) it increases fourth times Gravitational force is

(a) contact force

it remains the same

(b) non-contac* force (d) none of these

1. c 4. a 7. c 10 b

2 b 5 c 8. a 11; d

3. d e. c 9. c iar

. b 16. d 19. b 22.&2S. b 28. a . b 17. a 20. c 23. d -26. d 29. d

P. 5.1 Find the gravitational force of attraction between two spheres each of )

. . mass 1000 kg. The dJstt nee betwe«» the centres of the spheres is O.Snj. Given Data .’

Masses

Gravitational constant .ToFind: Gravitational force

‘Putting- the .valued

= 6.6%x icrtf

6.673

0.25

= 1000kg

/\ •.:,. G = 6.673 x 10″t1 Nm2 kf”2

F * ?

1000

N

UnH-S: Gravitation

\* q&^tt4gl

= 26.692 x F = 2.67 x «

Ans.

\

P.5.2 The gravitational force between two identical lead spheres kept at 1m apart is 0.006673 N, Find their masses.

Given Data: •;. Gravitational force Gravitational constant Distance between the masses

To Find: Masses mi = m2 – ? Calculation: Using the formula

F * 04J06673N

G – 6.673s x 10-11 Nm2 kg’2

;d = 1 m /’.•.’ ‘, “‘.-.-

(•.• Masses are identical)

F –

or

or

or

•= G

mxm

.'(v

F >-

m2 =

F x d2

Putting.the-values

m2 –

0.006673 N x (1m)2 6.673 JslO-1

m2 = 108kg2

m * 10,000 kg Ans. Thus, mass of each sphere is 10,000 kg.

Phy«lc8lX

P. 5.3 Find the acceleration due to gravity on the surface of the Mars. The mass of Mars is 6.42×1023 kg and its radius is 3370km.

Given Data . . .’ MassofMars –

-MU,, =6.42 x

f t*l -*

to Find

of Mars «’ Acceleration due to.gravityV

on the surface of mars Calculations: Using the formute

3370*m^33?0 k««DOO*m «*A3? x 10tm

‘ Lrt’. ‘ ‘ *.:>””<7.;>- ‘.’•”•’ ‘ .. ss. ?• . ‘ ‘ • . ‘ ‘.

or

mars

– Oi

mars .

Putting the values

gmars = 6.673 x 10-1

6.673 X6.42 Nm2 kg’

1Q71’x 1023

42.8 x 1012 x 12″12.. -i = 11.4 NKg

= 3.77 NKg-1

Since (1 N = 1kg ms’2), therefore , = 3.77 kg ms’2 x kg”1

gmars = 3.77 IDS”2 AnS.

P.5.4 The acceleration due to gravity on the surface of moon is 1.62 ms”2. The radiusof moon is 1740km. Find the mass of moon.

Given Data

‘Acceleration due to gravity”! * – 1 62 ~2 on the surface of moon ) , gmoon ^ i-p^ms

Radius of moon Rmoon = 1740km

x 1000m

To Find: Mass of moon Calculations: Using the formula

= ?

Unrt-5: Gravitation

Of , Qmoon =

moon

_., . >< . _..’ ymoon « rVmooo …

Of Mmoon = ^ q “ffra”-‘ •:••.’•• ‘. ‘?i’

Putting the values

M IMms”2 x (1.74- x 106>n)2 Mrtraon = 6.673 x 10’11 Nmzkg”z

1.62ms”2 x SxfO^W2′

6.673 x

4.86 x 1012 x 1011kg• 6.673; ‘•Vv- 5i /.

= 7.35 x 1022 kg AhS. /

P.5.5 Calculate the value of g at a height of 3600 km above the surface of the • _• earth, ”

Given Data

Height abqye earth’s surface h – 3600km – 3600 x 1000m

-Mass of earth Me – 6:0 x I024’k9

Radius of earth Re = 6.4x106m To Find: Gravitational acceleration g *= ? Calculations: Using the formula >!

or gh = G

Putting the values , •

gh = 6.673 x 10″” Nm2 kg”2 x

= 6.673 x 10-1 Nm2 kg-2 x = 6.673 x 10-“‘

100 x

tte

= 6.673 x 10:11×6.0x1012Nkg~

= 40.038 x, 10 kgms-2 x kg”1

– 4.0 ms~2 Ans.

P.5.6 Find the value of g due to the earth at geostationary satellite of the geostationary orbit is 48700 km. : w^ _

. The radius

Given Data

Radius of geostationary orbit

_ _. . /’Acceleration due to gravity^ 10 una: ^ at geostaUonary. s|i^lite 4 Calculations: Using the formula

g =

487QO km = 4870 x 1000 m f -•’.* 4.87 x 107m

or

Putting’the values

R

•_’ _„ , ._,, .. ,, -2 6;0 x 1024kg = 6.673 x 10 Nm2kg x.-(4^7;x1^7mp

– €,673 x 10-” Nm2kg-2 x

6^73 x GM* IP

“1

23-717 = 0.1688 kg ms:2 kg”1 ggeo = 0.17 ms~2 Arts. f ^ -, ,

P.5.7 The value of g is 4.0 ms”2 at a distance of 10000 km from the centre of , the earth. Find the mass of the earth. – h^

Given Data ;

Gravitational acceleration g = 4.0 ra^”4

Radius of earth Re -10000 knri« 10000×1000 m=107m

Calculation: Using the formula

Untt-5: Gravitation

..<£•

Putting the values

,>-2*

4.0 ms”2 x 101* m”

4.0×10

”

kg 1

Me

•” 6.673 = 0.599 x 1025 Jsg = 5.99X1024 kg

Ans.

P.5.8 At what altitude the value of g would become one fourth than on the , surface of the earth?

Given Data:

Mass of earth

M, * 610 x 10** kg

gh =

Acceleration due to

/gravity .-. ; ; ; . .. ‘., ;.( ^;}l,,^ , .;„.. , _•,,..; To Rnd: Altitude aix>v0eai^’s surface h =? ? Calculations: Using the formuJa

= 2.5 ms

~2

or (Rfh)2 = G^ Taking square root on both sidjBS

s

or

138

Physics IX

Putting ttie values

h

^

6.673 x 10~T1 Nm^g”2 x6.Q x 1CTkg : 2i5mS-2 ~;

-6.4x10em

= Vl60x101zmz – 6.4 x 106m = 12.65 x 100rn – 6.4 x 106 m

/ h H;£25 ^ 10sm ^

h * 6.4 x 106m h « (poeearth’sradius) Ans. ^

P.5.9 A polar satellite is laurtched at 850 km above earth. Find its orbital

speed.”

Given Data ‘:

Height h = 850 km – 850000 m = 0.85 x 106 m To Find: Orbital velocity v0 = ? . -Calculation: Using the formula

V0 =

As

Putting the values

(Re

Me

or

Putting the values

/6.673 x 10~1!Nm2 kg”2 x 6 x 10a kg v° = ^’ 6.4×10?m + o”85x106m

Unlt-5; Gravitation

‘40.038

7.25x10Bm

.038 x 10″ kg ms”^

40.038 x1013mzs” 7.25x10fim

7 „_-1

= >/5.5224x107ms

v0 = 7431 ms

-1

Ans.

P.5.10 A communication satellite is launched at 42000 km above earth. Find its orbital speed.

Given Data

Height h = 42000km = 42000x1000m = 42 x 106m To Find: Orbital velocity v0 = ?

“.'”…. ‘ ; ” # ‘ •

Calculation: Using the formula

v0 = V gh (R + h)

As gh =

Me

:, therefore

v0 =

or vQ =

Putting the values

/6.673x1Q-11Nm2kg-2 = V 6.4x106m + 423

x 10e m

140______;____________;P*iy«lc»tX

/40.038X lO

•V

48.4 x 10° m

40.038 x 10** kg

‘

40.038 .

484xlOB

> V0.8272 x 107 nr1 v0 = 2876ms’1 Ans.