Question 201: Gravitation

Question

A satellite of mass m is launched vertically upwards with an initial speed u from the surface of the earth. After it reaches height R (R = radius of the earth), it ejects a rocket
of mass $${m \over {10}}$$
so that subsequently the
satellite moves in a circular orbit. The kinetic energy of the rocket is (G is the gravitational constant; M is the mass of the earth) :

Accepted Answer

Using energy conservation
K
i
+ U
i
= K
f
+ U
f
$$ \Rightarrow $$ $${1 \over 2}m{u^2} - {{GmM} \over R}$$ = $${1 \over 2}m{v^2} - {{GmM} \over {2R}}$$
$$ \Rightarrow $$ v = $$\sqrt {{u^2} - {{GM} \over R}} $$
After ejecting a rocket of mass $${m \over {10}}$$ the remaining part of mass $${{9m} \over {10}}$$ will rotate the earth with orbital velocity v
0
.
$$ 	herefore $$ v
0
= $$\sqrt {{{GM} \over {2R}}} $$
Applying momentum conservation along radial direction,
Before firing rocket momentum of satelite in radial direction = mv
And after firing rocket momentum of satelite in radial direction = 0 and momentum of rocket in radial direction = $${m \over {10}}{v_2}$$
$$ 	herefore $$ mv = $${m \over {10}}{v_2}$$
$$ \Rightarrow $$ v
2
= 10v
Now applying momentum conservation along tangential direction we get,
0 = $${m \over {10}}{v_1}$$ - $${{9m} \over {10}}{v_0}$$
$$ \Rightarrow $$$${{9m} \over {10}}{v_0}$$ = $${m \over {10}}{v_1}$$
$$ \Rightarrow $$ v
1
= 9v
0
$$ 	herefore $$Total Kinetic Energy of rocket
= $${1 \over 2}{m \over {10}}\left( {v_1^2 + v_2^2} ight)$$
= $${1 \over 2}{m \over {10}}\left( {81v_0^2 + 100{v^2}} ight)$$
= $${m \over {20}}\left( {81\left( {{{GM} \over {2R}}} ight) + 100\left( {{u^2} - {{GM} \over R}} ight)} ight)$$
= $${m \over {20}}\left( {100{u^2} + {{81GM} \over {2R}} - {{100GM} \over R}} ight)$$
= $${m \over {20}}\left( {100{u^2} - {{119GM} \over {2R}}} ight)$$
= $$5m\left( {{u^2} - {{119GM} \over {200R}}} ight)$$

Answer

$${{3m} \over 8}{\left( {u + \sqrt {{{5GM} \over {6R}}} } \right)^2}$$

Answer

$${m \over {20}}\left( {{u^2} + {{113} \over {100}}{{GM} \over R}} \right)$$

Answer

$$5m\left( {{u^2} - {{119} \over {100}}{{GM} \over R}} \right)$$

Answer

$${m \over {20}}{\left( {u - \sqrt {{{2GM} \over {3R}}} } \right)^2}$$

Gravitation

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