A rocket is fired ‘vertically’ from the surface of mars with a speed of 2 km s–1. If 20% of its initial energy is lost due to Martian atmospheric resistance, how far will the rocket go from the surface of mars before returning to it? Mass of mars = 6.4× 1023 kg; radius of mars = 3395 km; G = 6.67× 10-11 N m2 kg–2. from Physics Gravitation Class 11 Manipur Board
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A rocket is fired ‘vertically’ from the surface of mars with a speed of 2 km s–1. If 20% of its initial energy is lost due to Martian atmospheric resistance, how far will the rocket go from the surface of mars before returning to it? Mass of mars = 6.4× 1023 kg; radius of mars = 3395 km; G = 6.67× 10-11 N m2 kg–2.

Initial velocity of the rocket, v = 2 km/s = 2 × 103 m/s

Mass of Mars, M = 6.4 × 1023 kg 

Radius of Mars, R = 3395 km = 3.395 × 106 m 

Universal gravitational constant, G = 6.67× 10–11 N m2 kg–2 

Mass of the rocket = 

Initial kinetic energy of the rocket = open parentheses 1 half close parenthesesmv

Initial potential energy of the rocket = fraction numerator negative space GMm over denominator straight R end fraction 

Total initial energy = open parentheses 1 half close parenthesesmv2 - fraction numerator negative space GMm over denominator straight R end fraction 

If 20 % of initial kinetic energy is lost due to Martian atmospheric resistance, then only 80 % of its kinetic energy helps in reaching a height. 

Total initial energy available = open parentheses 80 over 10 close parentheses x open parentheses 1 half close parentheses m v squared space minus space fraction numerator G M m over denominator R end fraction
= (80/100) × (1/2) 
mv2 - GMm over straight R 

                                        =  0.4
mv2 - GMm over straight R 

Maximum height reached by the rocket = 

At this height, the velocity and hence, the kinetic energy of the rocket will become zero. 

Total energy of the rocket at height h = fraction numerator GMm over denominator left parenthesis straight R plus straight h right parenthesis end fraction

Applying the law of conservation of energy for the rocket,

0.4mv2 - GMm over straight R  =  fraction numerator GMm over denominator left parenthesis straight R plus straight h right parenthesis end fraction

                 0.4v2 =GM over straight R space minus space fraction numerator G M over denominator left parenthesis R plus h right parenthesis end fraction 

                          = fraction numerator GMh over denominator straight R left parenthesis straight R plus straight h right parenthesis end fraction 

               fraction numerator straight R space plus straight h over denominator straight h end fraction space equals space fraction numerator GM over denominator 0.4 space straight v squared straight R end fraction  

                   straight R over straight h  = open parentheses fraction numerator GM over denominator 0.4 space straight v squared space straight R end fraction close parentheses space minus space 1

                      h = fraction numerator straight R over denominator open square brackets begin display style bevelled fraction numerator GM over denominator 0.4 space straight v squared straight R end fraction end style space minus space 1 space close square brackets end fraction 

                         = open parentheses fraction numerator 0.4 space straight R squared straight v squared over denominator GM space minus space 0.4 space straight v squared straight R end fraction close parentheses

=fraction numerator 0.4 straight x space left parenthesis 3.395 space straight x space 10 to the power of 6 right parenthesis squared space straight x space left parenthesis 2 space straight x 10 cubed right parenthesis squared over denominator left square bracket space 6.67 space straight x space 10 to the power of 11 space straight x space 6.4 space straight x space 10 to the power of 23 space minus space 0.4 space straight x space left parenthesis 2 space straight x 10 cubed right parenthesis squared space straight x space left parenthesis 3.395 space straight x space 10 to the power of 6 right parenthesis right square bracket end fraction 

  = fraction numerator 18.442 space straight x space 10 to the power of 18 over denominator left square bracket 42.688 space straight x space 10 to the power of 12 space minus space 5.432 space straight x space 10 to the power of 12 right square bracket end fraction

  = fraction numerator 18.422 space straight x space 10 to the power of 6 over denominator 37.256 end fraction 

  = 495 × 103 m

  = 495 km.
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The position co-ordinates of two particles of masses m1 and m2are (x1, y1, z1) and (x2, y2, z2) respectively. Find the coordinates of the centre of mass.

The position vectors of masses m1 and m2 are respectively,

space space space space space space space space space space space space r with rightwards harpoon with barb upwards on top subscript 1 equals space straight x subscript 1 straight i with overbrace on top plus straight y subscript 1 straight j with overbrace on top plus straight z subscript 1 straight k with overbrace on top

a n d space space space space space r with rightwards harpoon with barb upwards on top subscript 2 space equals space straight x subscript 2 straight i with overbrace on top plus straight y subscript 2 straight j with overbrace on top plus straight z subscript 2 straight k with overbrace on top

Let the position coordinates of the centre of mass be (X, Y, Z).

Therefore the position vector of centre of mass is,


R with rightwards harpoon with barb upwards on top equals straight X straight i with overbrace on top plus straight Y straight j with overbrace on top plus straight Z straight k with overbrace on top

Since,    R with rightwards harpoon with barb upwards on top equals fraction numerator straight m subscript 1 stack straight r subscript 1 with rightwards harpoon with barb upwards on top space plus straight m subscript 2 space straight r with rightwards harpoon with barb upwards on top subscript 2 over denominator straight m subscript 1 plus straight m subscript 2 end fraction space

therefore space space space straight X straight i with overbrace on top plus straight Y straight j with overbrace on top plus straight Z straight k with overbrace on top 

equals space space fraction numerator straight m subscript 1 left parenthesis straight x subscript 1 straight i with overbrace on top plus straight y subscript 1 straight j with overbrace on top plus straight z subscript 1 straight k with overbrace on top right parenthesis plus straight m subscript 2 left parenthesis straight x subscript 2 straight i with overbrace on top plus straight y subscript 2 straight j with overbrace on top plus straight z subscript 2 straight k with overbrace on top right parenthesis over denominator straight m subscript 1 plus straight m subscript 2 end fraction

space equals space space fraction numerator open parentheses straight m subscript 1 straight x subscript 1 plus straight m subscript 2 straight x subscript 2 close parentheses over denominator straight m subscript 1 plus straight m subscript 2 end fraction straight i with overbrace on top plus fraction numerator left parenthesis straight m subscript 1 straight y subscript 1 plus straight m subscript 2 straight y subscript 2 over denominator straight m subscript 1 plus straight m subscript 2 end fraction straight j with overbrace on topplus fraction numerator left parenthesis straight m subscript 1 straight z subscript 1 plus straight m subscript 2 straight z subscript 2 over denominator straight m subscript 1 plus straight m subscript 2 end fraction space straight k with overbrace on top

Comparing the coefficients of  straight i with overbrace on top comma space straight j with overbrace on top space and space straight k with overbrace on top, we get

straight X equals fraction numerator left parenthesis straight m subscript 1 straight x subscript 1 plus straight m subscript 2 straight x subscript 2 right parenthesis over denominator straight m subscript 1 space plus space straight m subscript 2 end fraction
straight Y space equals space straight X equals fraction numerator left parenthesis straight m subscript 1 straight y subscript 1 plus straight m subscript 2 straight y subscript 2 right parenthesis over denominator straight m subscript 1 space plus space straight m subscript 2 end fraction
straight z equals fraction numerator left parenthesis straight m subscript 1 straight z subscript 1 plus straight m subscript 2 straight z subscript 2 right parenthesis over denominator straight m subscript 1 space plus space straight m subscript 2 end fraction 


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