Gravitation Solve Numericals

Formulae

1. KEPLER'S THIRD LAW:

1.T² ∝ r³ (T square directly proportional to r cube)

Where,

T=the period of revolution of a body,

r=the radius of orbit in which the body is revolving.

2. (T1)² (r1)³

----- = -----

(T2) (r2)

Where T1,T2 and r1,r2 are periodic times and mean Radii of the orbit of two planets around the sun respectively.

2. GRAVITATIONAL FORCE BETWEEN TWO BODIES:

F=Gm1m2

-------------

r²

Where,m1 and m2 = masses of two bodies

r=distance of separation between them,

G=universal gravitational constant

3. UNIVERSAL GRAVITATIONAL CONSTANT:

G= Fr²

----------

m1m2

4. ACCELERATION DUE TO GRAVITY:

On the earth surface,g= GM

--------

R²

where, M=mass of the earth,

R=radius of the earth.

5. KINEMATICAL EQUATIONS OF MOTION:

1.v=u+at

2.s=ut+½at²

3.v²=u²+2as

FOR A FREELY FALLING BODY:

1.v=gt

2.h=½gt²

3.v²=u²+2as

FOR A BODY THROWN UPWARDS:

1.u=-gt

(Negative sign indicates velocity is decreasing)

2.h=ut-½gt²

3.u²=2gh

Where, u=initial velocity,

v=final velocity,

g=acceleration due to gravity,

h= distances of the body from the surfaces of the earth

6. POTENTIAL ENERGY OF A BODY:

1.On the Earth's surfaces, P.E.=-GMm

----------

2.At a height h from surfaces of earth,

P.E=-GMm

-----------

R + h

Where,m=mass of the body.

7. Escape velocity of a body(On the surface of the earth):

Vesc=√2GM

--------. = √2gR

VALUES TO REMEMBER

1.Gravitational constant(G)=6.7×10-¹¹ Nm²/kg²

2. Acceleration due to gravity(g)=9.8m/s²

3.Mass of the earth=6×10²⁴kg

4.Radius of the earth =6.4×10⁶m

Numerical section

TYPE 1 | KEPLER'S THIRD LAW

Formulae: T²∝r³

Let's the period of revolution of a planet at a distance R from a star be T, Prove that if it was at a distance of 2R from the star, it's period of revolution will be √8T.

Ans:

1. According to Kepler third law,the square of orbital period of revolution T of the planet around a star is directly proportional to the cube of the mean distance R of the planet from the star.

T²∝R³

Therefore,T²=k(R)³ ......(1)

Where, k is constant of proportionality.

2. When the planet is at a distance of 2R from the star, then it's period of revolution T' will be

T'²∝(2R)³

Therefore,T'²=k(2R)³ .......(2)

3.Dividing equations (1) and (2), we get,

T'² (R)³

---- = -------

T'² (2R)³

Therefore,T² 1

---- = ----

T'² 8

Therefore,T'= √8T

Thus, for a planet at a distance of 2R from the star , it's period of revolution will be √8T

TYPE 2 | GRAVITATIONAL FORCE BETWEEN TWO BODIES

Formulae :

1.F=Gm1m2

-------------

r²

2.G= Fr²

-----------

m1m2

1.Calculate the force between the sun and Jupiter. Assume that the mass of the Sun= 2 × 10³⁰kg, the mass of the Jupiter=1.89×10²⁷ kg and the radius of the jupiter orbit = 7.73 × 10¹¹m

(Use G = 6.67 × 10-¹¹Nm²/kg²)

Solution:

Given: mass of the Sun(m1)=2×10³⁰kg,

Mass of Jupiter(m2)=1.89×10²⁷

Radius of Jupiter orbit(r)

=7.73×10¹¹m,

Gravitational constant(G)

=6.67×10-¹¹Nm²/kg²

To Find : Force (F)

Formula: F = Gm1m2

--------------

r²

Calculation: From formula,

F=6.67×10-¹¹×2×10³⁰×1.89×10²⁷

---------------------------------------------

(7.73×10¹¹)²

= 6.67×3.78 25.21

----------------- ×10²⁴=--------×10²⁴

7.73×7.73 59.75

F=4.219×10²³N

Ans:The force between the sun and Jupiter is 4.219×10²³N

2. The masses of the earth and moon are 6 × 10²⁴ kg and 7.4 × 10²² kg, respectively.The distance between them is 3.84 × 10⁵ km. Calculate the gravitational force of attraction between the two.

Use G =6.7 × 10-¹¹ Nm²kg-²

Solution :

Given : Mass of the earth (Me) = 6 × 10²⁴ kg,

Mass of the moon (Mm) = 7.4 × 10²² kg,

Distance ( r ) = 3.84 × 10⁵ km

= 3.84 × 10⁸ m,

Gravitational constant (G)

= 6.7 × 10-¹¹Nm²/kg²

To find : Gravitational force ( F)

Formula : F = Gm1m2

--------------

r²

Calculation : From formula,

F= ( 6.7 × 10-¹¹) × (6×10²⁴)×(7.4×10²²)

--------------------------------------------------------

(3.84×10⁸)²

= 6.7 × 6 × 7.4

--------------------- ×. 10¹⁹

3.84 × 3.84

= 2 × 10²⁰

Ans : The gravitational force between the earth and the moon is 2 × 10²⁰N.

3. The mass of the earth is 6 × 10²⁴ kg. The distance between the earth and the sun is 1.5 × 10¹¹ m. If the gravitational force between the two is 3.5 × 10²²N, what is the mass of the Sun ? Use G = 6.7 × 10-¹¹ Nm²kg²

Solution :

Given : Mass of the earth (Me) = 6 × 10²⁴ kg,

Gravitational force (F) = 3.5 × 10²²N,

Distance (r) =1.5 × 10-¹¹m,

Gravitational constant (G)

= 6.7 × 10-¹¹ Nm²/kg²

To find : Mass of sun (Ms)

Formula : F = Gm1m2

--------------

r²

Calculation : From formula,

. Ms = fr²

-------

GMe

Therefore, Ms = ( 3.5 × 10²²)×(1.5×10¹¹)²

--------------------------------------

6.7 × 10-¹¹ × 6 × 10²⁴

= 7.88 × 10⁴⁴

-------------------=1.96×10³⁰kg

40.2×10¹³

Ans : The mass of the Sun is 1.96×10³⁰kg.

6.A spaceship is 10¹⁴ km away from a massive star. The gravitational force between them is 70N. Calculate the distance between them,if the force between them is increased to 7 × 10⁵N.

Solution :

Given : Force ( F1)=70N

Distance (r1)=10¹⁴km=10¹⁷m,

Force(F2)=7×10⁵N

To find : distance (r2)

Formula : F=Gm1m2

--------------

r²

Calculation : From formula,

For r1 = 10¹⁷m:F1=Gm1m2

-------------

(r1)²

70 = Gm1m2

---------------

(10¹⁷)² ......(1)

For r2:F2=Gm1m2

-------------

(r2)²

7 × 10⁵=Gm1m2

--------------

r2 ........(2)

Dividing equation (1) by (2),we have,

70 r²

--------- = ---------

7×10⁵ (10¹⁷)²

Therefore, r2² = 70

----------- × 10³⁴ = 10³⁰

7×10⁵

Therefore, r2 = 10¹⁵ m = 10¹² km

Ans: the distance between the spaceship and the star it's 10¹²km.

TYPE 3 | KINEMATICAL REQ EQUATIONS OF MOTION

Formulae :

1. Kinematical equation of motion :

1.v = u + at 2.s=ut+½at² 3.v² = u²+2as

2. For a freely falling body :

1.v = gt 2. h = ½ gt² 3. v²=2gh

3. For a body thrown upwards :

1. u = -gt(negative sign indicates velocity is decreasing)

2. h=ut-½gt²

3. u²=2gh

1. A metal ball of mass 5 kg fall from a height of 490 m. How much time it will take to reach the ground? (g=9.8m/s²)

Solution :

Given : mass (M)=5 kg, height (s)=490 m,

Gravitational acceleration (g)

=9.8 m/s²

To find : time taken (t)

Formula : s = ut + ½gt²

Calculation : from formula

490 = ( 0 × t ) + ( ½ × 9.8 × t²)

Therefore, 490= 4.9 t²

Therefore, t² = 490

---------- = 100

4.9

Therefore, t = 10s

Ans : the metal ball will take 10 s to reach the ground.

2. An object takes 5 s to reach the ground from a height of 5 m on a planet . What is the value of g on the planet?

Solution:

Given : Time (t( = 5 s , height (s)=5m

To find : Gravitational acceleration(g)

Formula : s = ut + ½gt²

Calculation : From formula,

5 = 0 × t + ½g(5)²

Therefore, 5 = ½g×25

Therefore, g=⅖

Therefore, g=0.4m/s²

Ans : the gravitational acceleration of the planet is 0.4m/s².

3.A stone is released from the top of a tower of height 19.6m. Calculate it's final velocity just before touching the ground.

Solution :

Given : Height ( h ) = 19.6m,

Initial velocity ( u ) = 0 m/s,

To find : Final velocity ( v )

Formula : v² = u² + 2as

Calculation : For downward motion,

a = g , s = h , u = 0

From formula,

v² = 2 × 9.8 × 19.6

Therefore, v² = 384.16

Therefore, v = 19.6m/s

Ans : Final velocity of the stone just before touching the ground is 19.6m/s.

4.A ball falls off a table and reaches the ground in 1 s. Assuming g=10 m/s², calculate it's speed on reaching the ground and the height of the table.

Solution :

Given : Initial velocity ( u ) = 0

Acceleration due to gravity(g)

=10m/s,

Time(t) = 1 s

To find : Height ( s ) , velocity ( v )

Formulae : 1. s = ut + ½ gt ²

2. v² = u² + 2 gs

Calculation : from formula (1)

s= 0 × t + ½ × 10 × 1² = 5m

From formula (2)

v² = 0+2×10×5

Therefore, v²=100

Therefore, v=10m/s

Ans: 1. The velocity of the ball when it reaches the ground is 10m/s

2. The height of the table is 5 m.

5. An object thrown vertically upwards reaches a height of 500 m. What was its initial velocity? How long will the object take to come back to the earth? Assume g = 10m/s²

Solution:

Given: Height ( s ) = 500m , acceleration due to gravity (g) = 10m/s²

To find: 1. Initial velocity ( u)

2. Total time taken

Formulae: 1.v²=u²+2as

2.s=ut+½at²

Calculation: For upward motion of the ball,(v)=0

a = -g = -10m/s²

From formula ( i )

0 = u² + 2 ( -10 ) × 500

u² = 10000

Therefore, u² = 100m/s

For downward motion of the ball,

(u) = 0 .

a = g = 10m/s²

From formula ( 2 )

500 = 0 +½ × 10t²

Therefore, t² = 500

-------- = 100

Therefore t = 10 s

Time for upward journey of the ball will be the same as time for downward journey i.e.,10 s.

Ans:1. The initial velocity of the object is 100m/s.

2. The total time taken by the object to reach the height and come down is 20 s

6.A tennis ball is thrown up and reaches a height of 4.05 m before coming down. What was its initial velocity? How much total time will it take to come down? Assume ji = 10m/s²

Solution:

Given: distance travelled by the ball ( s ) = 4.05 m, acceleration a=g=10m/s²

To find: 1. Initial velocity (u)

2. Time taken (t)

Formula: 1.v²=u²+2as

2.s=ut+½at²

Calculation: for upward motion of the ball, (v)=0.

a=-g=-10m/s²

Power formula (i)

0=u²+2(-10)×4.05

Therefore,u²=81

Therefore,u=9m/s

For downward motion of the ball

(u)=0.

a=g=10m/s²

For formula (2).

4.05=0+½×10t²

Therefore, t² = 4.05

--------- = 0.81

Therefore, t=0.9 s

Time for upward journey of the ball will be the same as time for downward journey i.e.,0.9 s.

Ans:1. The initial velocity of the ball is 9m/s

2. The total time taken by the ball to reach the ground is 1.8 s.

TYPE 4 | ACCELERATION DUE TO GRAVITY

FORMULA: on the earth surface, g=GM

------

R²

1. The mass and weight of an object on earth are 5 kg and 49 N respectively. What will be their values on the moon? Assume that the acceleration due to gravity on the moon is 1/6th of that on the earth.

Solution:

Given: Mass on the earth(me)=5kg, weight on earth(We)=49N,

Acceleration due to gravity on moon (gm)=9.8/6=1.63m/s²

To find: mass(mm), weight(Wm)on moon

Formula: Wm=mmgm

Calculation: the mass of the object is independent of gravity and remains unchanged i.e 5 kg.

From formula,

Wm=5×1.63

Therefore, Wm=8.15N

Ans: On moon the mass of the object is 5 kg and weight is 8.15 N.

2. The radius of planet A is half the radius of planet B. If the mass of A is Ma, What must be the mass of B so that the value of g on B is half that of it's value on A?

Solution:

Given: For planet A : mass = Ma,

radius Ra=Rn

------

Acceleration due to gravity=gA

For planet B:radius =Rb,

Acceleration due to gravity gb=gA

-----

To find: mass of planet B(Mb)

Formula : g=GM

--------

R²

Calculation: from formula

Ma gaRa² G

------ = ---------- × -------

Mb G gbRb²

........(substituting for Ra and gb)

Therefore, Ma 1

-------- = ------

Mb 2

Therefore, Mb=2Ma

Ans: the mass of planet B is 2Ma.

TYPE 5 | ESCAPE VELOCITY

FORMULA : Vesc=√2GM

-------- = √2gR

1. Calculate the escape velocity on the surface of the moon given the mass and the radius of the moon to be 7.34 × 10²² kg and 1.74 × 10⁶ m respectively.

Solution:

Given: gravitational constant ( G ) = 6.67 × 10-¹¹ Nm²/kg², mass of the moon ( R ) = 1.74 × 10⁶m.

To find: Escape velocity ( Vesc )

Formula: Vesc=√2GM

---------

Calculation: From formula,

Vesc=√2×6.67×10-¹¹×7.34×10²²

-------------------------------------

1.74 × 10⁶

=√97.9×10⁵

--------------

1.74

Therefore, Vesc=2372m/s=2.372km/s

Ans: Escape velocity on the moon is 2.372km/s.

2.What would be the duration of the year if the distance between the earth and sun gets doubled? Will this affect the escape velocity of an object on the surface of the earth?

Solution:

Given: Distance(r')=2r

Time period(T)=365 days

To find: time period (T)

Formula: T²∝r³

Calculation: From formula,

T² r³ r³

--- = ------ = ------

T² r'³ 8r³

Therefore, T'=√8T = √8(365)

Therefore, T ' = 1032.37 days

For an object on the surface of the earth, it's escape velocity is independent of radius of orbit, hence this will not affect its escape velocity.

I hope you understand in easy sentences

Gravitation Solve Numericals | Class 10th | Maharashtra Board