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Minimum energy required to launch

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The minimum energy required to launch a m kg satellite from earths surface in a circular orbit at an atitude of 2R , where R is the radius of earth .

The minimum energy should be (1/2) mv$^2$
As it is circular orbit
$$mv^2/(3R) = GMm/(3R)^2$

Solving this I got
$$(1/2)mv^2 = GMm/(6R) = mgR/6$

But the answer is given as (5/6)mgR

asked Feb 25, 2017 in Physics Problems by koolman (4,116 points)
You have only found the KE in the orbit. The question is asking for the KE at the launch point.

Try applying Conservation of Energy.
Do I have to consider PE at ground
You have to consider the difference in PE between ground and orbit :  
KE+PE at ground = KE+PE in orbit.
Now what is KE at ground.
That is what the question is asking you to find.
So it should be
$$KE_s = KE_o + PE_o - PE_s$$
$$KE_s = mgR/6 -mgR/3 + mgR$$
$$KE_s = 5mgR/6$$

1 Answer

2 votes
 
Best answer

Energy is conserved :
KE0+PE0 at launch = KE+PE in orbit.

The orbit is circular and has radius 3R so
$mv^2/3R=GMm/(3R)^2$
$KE=\frac12mv^2=GMm/6R$.

The energy equation is
$KE_0-GMm/R=KE-GMm/3R$
$KE_0=GMm/R+GMm/6R-GMm/3R=(GMm/R )(\frac66+\frac16-\frac26)=\frac56\frac{GMm}{R}$.

At the surface of the Earth
$mg=GMm/R^2$ so $GMm/R=mgR$.
Therefore the energy required for launch is at least
$KE_0=\frac56 mgR$.

answered Feb 25, 2017 by sammy gerbil (26,096 points)
selected Feb 26, 2017 by koolman
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