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A Saturn year is 29.5 times the earth year. How far is the Saturn from the sun if the earth is 1.50 x 1011m away from the sun?

According to Kepler’s third law of planetary motion,

                straight T squared proportional to straight R cubed 

∴       open parentheses straight R subscript straight s over straight R subscript straight e close parentheses cubed space equals space open parentheses straight T subscript straight s over straight T subscript straight e close parentheses squared 

rightwards double arrow            straight R subscript straight s equals straight R subscript straight e open parentheses straight T subscript straight s over straight T subscript straight e close parentheses to the power of 2 divided by 3 end exponent                  ... (1)

Given that, Saturn year is 29.5 times the earth year.

i.e., Ts = 29.56 TE

Distance of Earth from Sun = 1.50 x 1011 m

Therefore, 
                   Rsequals 1.5 cross times 10 to the power of 11 left parenthesis 29.56 right parenthesis to the power of 2 divided by 3 end exponent

                       equals 1.43 cross times 10 to the power of 12 straight m 
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The escape speed of a projectile on the earth’s surface is 11.2 km s–1. A body is projected out with thrice this speed. What is the speed of the body far away from the earth? Ignore the presence of the sun and other planets. 


Escape velocity of a projectile from the Earth, vesc = 11.2 km/s 

Projection velocity of the projectile, vp = 3vesc 

Mass of the projectile = 

Velocity of the projectile far away from the Earth = v

Total energy of the projectile on the Earth = mvp2 - mvesc

Gravitational potential energy of the projectile far away from the Earth is zero. 

Total energy of the projectile far away from the Earth = mvf

From the law of conservation of energy, we have 

mvp2 - mvesc2  =   mvf
                               vf = ( vp2 - vesc)1/2 

                                  = [ (3vesc)2 - vesc2 ]1/2 

                                  = √8 vesc 

                                  = √8 × 11.2 

                                  =  31.68 km/s.
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A Saturn year is 29.5 times the earth year. How far is the saturn from the sun if the earth is 1.5 x 108 km away from the Sun?

According to Kepler’s third law of planetary motion,
                             straight T squared proportional to straight R cubed
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Here,           straight T subscript straight s equals 29.5 straight T subscript straight e   and    straight R subscript straight e equals 1.5 cross times 10 to the power of 8 km

∴          straight R subscript straight s space equals space straight R subscript straight e open parentheses straight T subscript straight s over straight T subscript straight e close parentheses to the power of 2 divided by 3 end exponent
               space space space equals 1.50 cross times 10 to the power of 8 left parenthesis 29.5 right parenthesis to the power of 2 divided by 3 end exponent space km
                   equals 1.432 space cross times space 10 to the power of 12 km
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A satellite orbits the earth at a height of 400 km above the surface. How much energy must be expended to the satellite to rocket out of the earth’s gravitational influence?
Mass of the satellite = 200 kg, Mass of the earth = 6.0 x 1024 kg, Radius of the earth = 6.4 x l06 m; G = 6.67 x 10-11 Nm2/kg2.


Kinetic energy of the satellite is, 

1 half mv subscript straight o squared space equals space 1 half straight m open parentheses fraction numerator GM over denominator straight R plus straight h end fraction close parentheses 

            equals 1 half cross times 200 open parentheses fraction numerator 6.67 cross times 10 to the power of negative 11 end exponent cross times 6.0 cross times 10 to the power of 24 over denominator 6.8 cross times 10 to the power of 6 end fraction close parentheses

            equals space 5.89 space cross times space 10 to the power of 9 space straight J 

Potential energy of the satellite is, 

space space space space minus fraction numerator GMm over denominator straight R plus straight h end fraction equals negative open parentheses fraction numerator 6.67 cross times 10 to the power of negative 11 end exponent cross times 6.0 cross times 10 to the power of 24 cross times 200 over denominator 6.8 cross times 10 to the power of 6 end fraction close parentheses 

               equals negative 11.78 space cross times space 10 to the power of 9 straight J 

Total energy of satellite is,

E = 5.89 x 109 - 11.78 x l09

   =-5.89 x 109

The total energy of the satellite is zero, if the satellite just escapes from the gravitational field. Therefore, 5.89 x 109J of energy has to supplied to the satellite to escape the gravitational pull.
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A spaceship is stationed on Mars. How much energy must be expended on the spaceship to launch it out of the solar system? Mass of the space ship = 1000 kg; mass of the Sun = 2 × 1030 kg; mass of mars = 6.4 × 1023 kg; radius of mars = 3395 km; radius of the orbit of mars = 2.28 × 108kg; G= 6.67 × 10–11 m2kg–2.

Mass of the spaceship, m= 1000 kg

Mass of the Sun, M = 2 × 1030 kg 

Mass of Mars, mm = 6.4 × 10 23 kg 

Orbital radius of Mars, R = 2.28 × 10kg

                                    =2.28 × 10
11

Radius of Mars, = 3395 km

                          = 3.395 × 10
6 m 

Universal gravitational constant, G = 6.67 × 10–11 m2kg–2 

Potential energy of the spaceship due to the gravitational attraction of the Sun =

Potential energy of the spaceship due to the gravitational attraction of Mars = - 

Since the spaceship is stationed on Mars, its velocity and hence, its kinetic energy will be zero. 

Total energy of the spaceship = 
                                            =  

The negative sign indicates that the system is in bound state. 

Energy required for launching the spaceship out of the solar system, 

= – (Total energy of the spaceship) 

=  

= 6.67 × 1011 × 103 ×

596.97 × 109 

=  6 × 10
11 J.
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