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Tuesday, June 30, 2020

Force numericals SEE

SG

 Unit-1                                           Force               ( class:-10)

 

     NUMEICAL SOLUTION

1.     The distance between earth and moon is 36x108m. If the mass of the earth is 6x1024kg and mass of the moon is 7.2x1022kg. What is the gravitational force between them?

Solution:-

Given:-

Distance between them(d) = 36x108m = 3.6x109m

Mass of the earth (M) = 6x1024kg

Mass of the moon (m) = 7.2x1022kg

Gravitational force (F) = ?

Gravitational constant (G) = 6.67 x10-11 Nm2/kg2.

Now, By using formula

 F = GMm/d2 =6.67 x10-11 x 6x1024 x 7.2x1022 /(3.6x109)2

Or , F = 288.144 x 1035 / 12.96 x1018 = 22.23 x 1017

\ F = 2.223 x 1018 N

2.     The mass and radius of Jupiter is 1.9x1027 kg and 71x106m respectively. Find the acceleration due to gravity of it. What will be the weight  of an object having mass 75 kg on that planet?

Solution:-

Given:-

Mass of the Jupiter (M) = 1.9x1027

Mass of an object (m) = 75 kg

Radius of the Jupiter( R ) =71x106m = 7.1x107m

Acceleration due to gravity on Jupiter surface (g) = ?

Weight of an object (w) = ?

Now , By using formula

  g = GM/R2 /

or, g = 6.67 x10-11 x 1.9x1027  / (7.1x107 )2

or, g = 12.67 x1016 / 50.41 x 1014

or, g= 0.2513 x 102

\ g = 25.13 m/s2

Again,

\W = mxg = 75 x 25.13 = 1884.75 N

3.     The mass of Jupiter is 319 times greater than the mass of earth and the radius is 11 times greater than radius of earth. If the acceleration due to gravity on the earth surface is 9.8 m/s2. Calculate the acceleration due to gravity on the surface of Jupiter.

Solution:-

Given:-

Mass of the Jupiter (MJ) = 319 Me.

Radius of the Jupiter(RJ) = 11 Re.

Acceleration due to gravity on the earth surface (ge) = 9.8 m/s2.

Acceleration due to gravity on the Jupiter surface (gj) = ?

Now,

Acceleration due to gravity on the earth surface is

  g=  GMe / Re2  or   9.8 =  GMe / Re2  …………………..eqn (i)

Acceleration due to gravity on the earth surface is

     g=  GMj / Rj2 

or, gj = G 319 Me / (11Re)2

or, gj = 319 GMe / 121Re2

or, gj = 319 / 121 x 9.8                                       [ GMe / Re2  = 9.8]

\ gj  = 25.83 m/s2.

4.     The earth is compressed to the size of the moon. Calculate the acceleration due to gravity of newly formed earth based on the data given below:-

Mass of the earth (Me) = 6x1024 kg

Mass of the moon (Mm) = 7.2x1022 kg.

Radius of the earth (Re) = 63800 km

Radius of the moon (Rm) = 1.7x103km

Solution:-

Given:-

Mass of the earth (Me) = 6x1024 kg

Radius of the moon (Rm) = 1.7x103km =1.7x106m

Acceleration due to gravity of new earth(gn) = ?

gn = GMe / Re2  = 6.67x10-11 x 6x1024 / (1.7x106)2 = 40.02 x 1013/ 2.89x1012

  or, gn = 13.847x10

\ gn  = 138.47  m/s2.

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