Question

✪ Physics ✪
The energy required to take a satellite to a height 'h' above earth surface (radius of earth = 6.4 × 10³ km) is E1 and kinetic energy required for the satellite to be in a circular orbit at this height is E2. The value of h for which E1 and E2 are equal is :-
a) 1.28 × 10⁴ km
b) 6.4 × 10³ km
c) 3.2 × 10³ km
d) 1.6 × 10³ km​

Answer 1

★Given :

• The energy required to take a satellite to a height 'h' above earth surface is E₁.
• Kinetic energy required for the satellite to be in a circular orbit at this height is E₂ .
• Radius of earth = 6.4 × 10³ km

★To find :

• The value of h for which E₁ and E₂ are equal

★Solution :

Applying conservation of total energy,

(K.E + P.E)₁ = (K.E + P.E)₂ __(1)

K.E of satellite is zero at the earth's surface.

And if we define the potential energy to be zero at r = infinity,U = -GmM/r.

Therefore,

$\displaystyle \implies \sf E_1+\bigg(\frac{-GMm}{R} \bigg)= 0+\bigg(\frac{-GMm}{(R+h)} \bigg)$

$\displaystyle \implies \sf E_1 = \frac{GMm}{R} - \frac{GMm}{(R+h)}$

$\displaystyle \implies \sf E_1 = \frac{GMm(R+h)-GMm(R)}{R(R+h)}$

$\displaystyle \implies \sf E_1 = \frac{GMm(\cancel{R}+h-\cancel{R})}{R(R+h)}$

$\displaystyle \implies \sf E_1 = \frac{GMmh}{R(R+h)}$

We know :

$\mapsto \boxed{\sf V =\sqrt{\frac{GM}{r}}}$

$\therefore \sf V = \dfrac{GM}{R+h}$

Now,

Using the formula :

$\leadsto \boxed{\sf K.E = \frac{1}{2} mv^2 }$

$\implies \sf E_2 = \dfrac{1}{2} mv^2$

$\implies \sf E_2 = \dfrac{1}{2} \times m\times \bigg(\sqrt{\dfrac{GM}{(R+h)}} \bigg) ^2$

$\implies \sf E_2 = \dfrac{1}{2} \times m \times \dfrac{GM}{R+h}$

We have to find the value of h for which E₁ & E₂ are equal.Therefore,

$\implies \sf E_1 = E_2$

$\displaystyle \implies \sf \frac{\cancel{GMm}h}{R\cancel{(R+h)}} = \frac{1}{2} \cancel{m}\bigg(\frac{\cancel{GM}}{\cancel{R+h}} \bigg)$

$\implies \sf \dfrac{h}{R} = \dfrac{1}{2}$

$\implies \sf 2h = R\\\\ \implies \sf h = \dfrac{R}{2}$

Given that R = 6.4 × 10³ km.

Substituting this value in the above equation,

$\implies \sf h = \dfrac{6.4 \times 10^3 }{2}$

$\implies \underline{\boxed{\bold{h = 3.2\times 10^3km}}}$

Therefore,

Option(C) is correct.

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