Acceleration due to gravity | Definition , Equation , Variation and Solved examples

Acceleration due to Gravity

When we dropped a body from a certain height above the ground it begins to fall towards the earth due to the gravity . The acceleration generated in the body due to gravity is called the acceleration due to gravity . It is generally denoted by g . Its value near or on the surface of earth is 9.8 m/s².

Suppose , the mass of the Earth is M and it’s radius is R , the force of gravitation attraction acting on a body of mass m close to the surface of earth is

F = GMm/R²

According to Newton’s second law , the acceleration due to gravity

g = F/m = GM/R²

The value of g is independent of the shape, size, mass, etc. of the body but depends upon mass and radius of the earth or planet due to which there is a gravity .

When two bodies of different masses are allowed to fall from same height in vacuum, they reach the earth at the same time. If the two bodies of same mass but of different volumes are allowed to fall from same height in air medium, they will not be reaching simultaneously. The body of smaller volume will reach the earth earlier than the other body, since the upward thrust due to air on the body of smaller volume will be less than that on the body of larger volume .

Relation between ‘g’ and ‘G’

Let’s Consider earth is a spherical body of mass M and radius R , center is denoted by the symbol O . Suppose a body is placed on the surface of eatrth that have mass ‘ m ‘ . The force with which a object is attracted towards the centre of earth due to gravitational atteraction is called the weight of the body .

So , the force with which the body will be attracted is

F= GMm/R² …( i )

where M=mass of the earth and R = radius of the earth.

If g denotes the acceleration due to gravity , then the force acting on the body due to earth will be

Force = mass x acceleration

F = mg … ( ii )

From (i) and (ii), we have

mg = GMm/R²

g = GM/R²

or g = [ (4/3)πρR³]G/R³

g = (4/3)πρGR

Value of ‘g’ on Earth’s surface 

Earth is a like a solid sphere , For a point on the surface or out side of surface we can assume all the masses concentrate on surface of earth’s center .

So , for earth value of g on earth surface is

gs = GMe/Re²

Variation in Value of ‘g’ due to shape of earth

Till now we’ve considered that earth is spherical in its shape but this is not actually true. Due to some geological and astronomical reasons, the shape of earth is not exact spherical. It is ellipsoidal as shown in figure 22.1 

Fig. 22.1

As we’ve discussed that the value of g at a point on earth surface depends on radius of Earth. As we can see from figure 22.1 that at poles radius of Earth is small compared to that at equatorial points. It is observed that the approximate difference in earth’s radius at different points on equator and poles is re-rp ≈ 21 to 34 km. Due to this the difference in value of g at poles and equatorial points is approximately gp-ge ≈ 0.02 to 0.04 m/s², which is very small. So for numerical calculations, generally, we ignore this factor while taking the value g and we assume Earth spherical in shape .

Variation in value of ‘g’ with height from earth surface 

Let’s Assume a body of mass m is at height ‘h’ from earth surface . 

Fig. 22.2

The force of gravity on object is

F= GMm/(R+ h)²

Acceleration due to gravity at this height will be ,

g’=F/m = GM/(R+h)²

This can also be written as,

g’ = GM/[R²(1+(R/h))²]

we know that g = GM/R²
So ,
g’ = g/ [1+ (h/R)]²

Thus g'< g
ie .The value of acceleration due to gravity g goes on decreasing as we go above the surface of earth .
Further,
g’ = g [1 + (h/R)]-2

or g’ ≈ g [1- ( 2h/R)] if h << R

Variation in value of ‘g’ with depth from earth surface

Suppose Earth to be a homogenous sphere
(i.e., a sphere of uniform density).

Let ρ denotes the mean density of the Earth

M = mass of the Earth (supposed concentrated at its centre)

R = radius of the earth

g= acceleration due to gravity at the point P on the surface of the Earth

d = depth of a point P” inside the Earth and gd = acceleration due to gravity at the point P .
Clearly,
g= GM/R² = [G×(4π/3)R³ρ]/R²

g = 4πGRρ/3. …( i )

Fig. 22.3

With O as centre and (R- d) as radius, let us draw an imaginary sphere so that Earth is divided into two parts (refer figure 22.3 ).

(i) a shell of thickness d denoted by I.
(ii) a sphere of radius (R- d) denoted by II.

The force of attraction on a body placed at the point P” is due to:
(i) shell I and (ii) sphere II.

We know that a body lying inside a shell does not experience any force. Hence, the only force with which the body at P” is attracted is due to the sphere II.

Thus, gd= acceleration due to gravity at P” due to sphere II (as the shell I does not contribute in this respect).

Assuming the density ρ to be constant throughout the Earth,
we can replace R by (R-d) in Eq. (i) and get

gd= 4πG(R-d)ρ/3 …..( ii )

Dividing Eq. (ii) by Eq. (i), we get

gd = g[1-(d/R)]

Clearly, the acceleration due to gravity decreases as we move down into the surface of the Earth, i.e., acceleration due to gravity in the mines is less than that on the surface of the Earth .

At the centre O of the Earth , d = R Thus, gd = 0

Hence, acceleration due to gravity at the centre of the Earth is zero.

Fig. 22.4 Graph of variation in the value of g with r (the distance from center of earth )

Effect of Earth’s Rotation on Value of ‘g’

Fig. 22.5

Let us consider a body of mass m placed on Earth’s surface at a latitude λ as shown in figure 22.5  This mass experiences a force mgs ,  towards the centre of earth and a centrifugal force mωe2Re2Sinλ relative to Earth’s surface as shown in figure. If Nis the normal contact force on mass then for equilibrium of body we have

N + mω2ReSin2λ=mgs
N = mgs – mω2ReSin2λ

Here we can see that the normal contact force on body is less

then mgs on Earth’s surface. At a point on Earth surface we don’t feel this centrifugal force but actually it acts on us and due to this the effective weight of mass is decreased. If we consider geff as the effective value of g on earth surface at a latitude λ then we can write

mgeff=mgs – mωe²ReSin²λ

geff=gs – ωe²ReSin²λ. …. ( i )

From equation- ( i ) we can find the value of effective gravity at poles and equatorial points on Earth as

At poles λ = 0
gpoles =gs = 9.83 m/s²

At equator λ = π/2
gequator = gs – ω²R = 9.78 m/s
²

Thus we can see that if the body is placed at poles of Earth , it will have only a spin , not circular motion ,That’s why there is no reduction in value of g due to rotation of earth at poles . Thus at poles value of g on Earth surface is maximum and at equator it is minimum . But an average we take 9.8 m/s² , the value of g everywhere on earth’s surface .

Solved Examples

  • Problem 1 .The acceleration due to gravity at the Moon’s surface is 1.67 m/s² . If the radius of the Moon is 1.74 x 10⁶ m, calculate the mass of the Moon. ( If the value of G is 6.67×10-11 Nm²/kg² .

Solution. We know that
g= GM/R²

or M = gR²/G

Here, g =1.67 m/s²
R=1.74×10⁶ m
G = 6.67×10-11 Nm²/kg²

Thus,
M = (1.67)(1.74×10⁶)²/6.67×10-11

M = 7.6×10²² Kg

  • Problem 2 . Calculate the value of acceleration due to gravity at a point given below .
    (a) 5:0 km above the earth’s surface and
    (b) 5:0 km below the earth’s surface.

Radius of earth 6400 km and the value of g at the surface of the earth is 9.80 m/s²

Solution – (a) The value of g at a height h is (for h<<R)

g=go[1-(2h/R)]

= ( 9.80m/s² )[1-( 2×5.0 km)/6400 km]

= 9.78 m/s²

(b) The value at a depth h is

g = go[1-( h/R )]

= ( 9.80m/s² )[1-( 5.0 km/6400 km) ]

= 9.79 m/s²

  • Problem 3. At what height from the surface of earth the acceleration due to gravity reduce by 19% of its value on the surface of earth ? ( given that
    Radius of earth = 6400 km )

Solution- Since the acceleration due to gravity reduces by 19%, the value of acceleration due to gravity there is 100 – 19 = 81%. It
means, g’ = 81g/100. If h is the height of location above the surface of earth, then

g’= gR²/(R+h)²

or 81g/100 = gR²/( R+h )²

9/10 = R/(R+h)

9R+ 9h = 10R

h=R/9 = R/9 = (6.4×10⁶)/9

h = 7.111×105 m

  • Problem 4 – At what depth below the surface of earth, value of acceleration due to gravity is same as the value at height h=R , where R is the radius of earth

Solution : Let’s the value of g at depth ‘d’ is same as the value of g at height R from earth surface.
then

g[1-(d/R)] = g/[1+( h/R)]²

put the value of d = R
we get ,

d = 3R/4

So , we can say that value of g at depth 3R/4 is same as at height R from earth surface .

  • Problem 5 . If by maintaining the same mass as before , the radius of the earth is shrink by two percentage then What would happen to the acceleration due to gravity on the surface of earth ?

Solution – Consider the case of a body of mass m placed on the earth’s surface (mass of the earth M and radius R ). and g present the acceleration due to gravity, then we know that

gs= GMe/Re² ….( i )

Now, when the radius is reduced by 2 %, i.e., radius becomes 0.98 R, let acceleration due to gravity be g’, then

g’= GM/(0.98R)² ….( ii )

With help of equation ( i ) and ( ii ), we get

g’/g = R²/(0.98R)² = 1/(0.98)²

g’ = g×(1/0.98)²

g’= 1.041 g

Thus, the value of g is increased by 4.1 %.

  • Problem 6 -Determine the speed with which the earth would have to rotate on its axis, so that a person on the equator would weight (3/5)th as much as at present .
    Take R = 6400 km.

Solution- g’= 3g/5

g’= g – Rω²Cos²0°

Rω²= 2g/5
or
ω = [ 2g/5R ]1/2

ω = [ (2×9.8)/(5×6400×10³)]

ω = 7.8 × 10-4 rad/s

  • Problem 7 – Suppose the earth increases its speed of rotation. At what new time period will the weight of a body on the equator becomes zero?
    Take g = 10 m/s² and radius of earth R=6400 km.

Solution-
The weight will become zero , when
g’=0 or g-Rω² =0

ω = ( g/R )1/2

2π/T = ( g/R )1/2

or T = 2π( R/g)1/2

Substituting the values,

T = [ 2π ( 6400×10³ /10 )1/2 ]/3600

T = 1.4 h

Thus, the new time period should be 1.4 h instead of 24 h for the weight of a body to be zero on the equator .

  • Problem 8 – Assuming earth as a uniform sphere , Find the percentage increase in the angular velocity of earth , so that , all bodies lying on the equator, just leave the surface of earth .

Solution – Let us assume that the acceleration due to gravity, at present, at the equator as

9.8 = GM/R² – Rωo² …( i )

where ωo is the present value of angular velocity of earth .

Now, if ω be the new angular velocity when, the bodies at the
equator leave the earth’s surface then ge = 0

Now, if   ω be the new angular velocity when, the bodies at the equator leave the earth’s surface then ge = 0

0 = GM/R² – Rω²

GM/R² = Rω² ….. ( ii )

with the help of equation ( i ) and ( ii )

9.8 = Rω² – Rωo²

ω² – ωo² = 9.8/R
ω² = ωo² + 9.8/R

Now % increase in ω
= (ω – ωo)×100/ωo
= [ (ω/ωo)-1] ×100

= [ [( 9.8/Rωo²)+1 ]1/2 -1]×100

Put the value
ωo = (2π/ 86400) rad/sec
and R = 64× 10⁵ m

The value comes out to be nearly 1700%

i.e. the angular velocity of earth should increase about 17 times .

  • Problem 9 . At a depth h1=R/2 from the surface of the earth acceleration due to gravity is g1 ,  Its value changes by △g1 , when one moves down further by 1 km. At a height h2 above the surface of the earth acceleration due to gravity is g2 .  Its value changes by △g2 , when one moves up further by 1 km. If △g1 = △g2 find h2 .  Consider earth as a uniform sphere of radius R .

Solution- Acceleration inside the Earth at a distance r from the centre

g1 = GMr/R³
= GM(R-h1)/R³
= g[1-(h1/R)]

△g1= ( dg1/dh1)△h1 = – g△h1/R
△h1 = +1 km
△g1 = – (g/R)(1km) …..( i )

( This is independent of depth h1 )

At a distance r from the center outside the earth

g2 = GM/r²
= GMR²/R²r²
= gR²/(R+h2

△g2 = (dg2/dg1)(△h2)
= -2gR²△h2/(R+h2

△h2 = 1 km

△g2 = – 2gR²(1km )/(R+h2)³ … ( ii )

Since , △g1 = △g2

(1 km)g/R = – 2gR²(1 km)/(R+h2

( R + h2 )³= 2R³

1+( h2/R ) = ( 2 )1/3

h2 = R[21/3 – 1]

Leave a Comment

Your email address will not be published.