Question

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:
1. 1.6 × 10³ km
2. 3.2 × 10³ km
3. 6.4 × 10³ km
4. 1.28 × 10⁴ km

Answer 1

Correct Answer - Option 2 : 3.2 × 10³ km

Concept:

The energy required to take a satellite to a height 'h' above earth’s surface is difference between the energy at height ‘h’ and energy at surface, then,

⇒ E₁ = U_h - U_i

Since we need to find the energy at that point, the gravitational potential energy is used here.

The gravitational potential energy is given by the formula:

\({\rm{U}} = - \frac{{{\rm{GMm}}}}{{\rm{r}}}\)

The energy at height ‘h’ is given by the formula:

\({{\rm{U}}_{\rm{h}}} = - \frac{{{\rm{G}}{{\rm{M}}_{\rm{e}}}{\rm{m}}}}{{{{\rm{R}}_{\rm{e}}} + {\rm{h}}}}\)

The energy at surface is given by the formula:

\({{\rm{U}}_{\rm{i}}} = - \frac{{{\rm{G}}{{\rm{M}}_{\rm{e}}}{\rm{m}}}}{{{{\rm{R}}_{\rm{e}}}}}\)

On substituting the values of energy in above energy equation,

\(\Rightarrow {{\rm{E}}_1} = - \frac{{{\rm{G}}{{\rm{M}}_{\rm{e}}}{\rm{m}}}}{{{{\rm{R}}_{\rm{e}}} + {\rm{h}}}} + \frac{{{\rm{G}}{{\rm{M}}_{\rm{e}}}{\rm{m}}}}{{{{\rm{R}}_{\rm{e}}}}}\)

The kinetic energy (E₂) required for the satellite to be in a circular orbit at this height is given by the formula:

\({{\rm{E}}_2} = \frac{1}{2}{\rm{m}}{{\rm{v}}^2}\)

Where,

m = Mass of the satellite

v = Velocity of the satellite while it is moving in circular orbit

The velocity of the satellite when it is orbiting is given by the formula:

\({\rm{v}} = \sqrt {\frac{{{\rm{G}}{{\rm{M}}_{\rm{e}}}}}{{{{\rm{R}}_{\rm{e}}} + {\rm{h}}}}} \)

On substituting velocity in kinetic energy in the equation,

\({{\rm{E}}_2} = \frac{{\rm{m}}}{2}\left( {\frac{{{\rm{G}}{{\rm{M}}_{\rm{e}}}}}{{{{\rm{R}}_{\rm{e}}} + {\rm{h}}}}} \right)\)

Calculation:

From question, we need to find the height where both the energies E₁ and E₂ are equal.

\(\Rightarrow - \frac{{{\rm{G}}{{\rm{M}}_{\rm{e}}}{\rm{m}}}}{{{{\rm{R}}_{\rm{e}}} + {\rm{h}}}} + \frac{{{\rm{G}}{{\rm{M}}_{\rm{e}}}{\rm{m}}}}{{{{\rm{R}}_{\rm{e}}}}} = \frac{{\rm{m}}}{2}\left( {\frac{{{\rm{G}}{{\rm{M}}_{\rm{e}}}}}{{{{\rm{R}}_{\rm{e}}} + {\rm{h}}}}} \right)\)

\(\Rightarrow {\rm{G}}{{\rm{M}}_{\rm{e}}}{\rm{m}}\left[ { - \frac{1}{{{{\rm{R}}_{\rm{e}}} + {\rm{h}}}} + \frac{1}{{{{\rm{R}}_{\rm{e}}}}}} \right] = {\rm{G}}{{\rm{M}}_{\rm{e}}}{\rm{m}}\left[ {\frac{1}{{2\left( {{{\rm{R}}_{\rm{e}}} + {\rm{h}}} \right)}}} \right]\)

\(\Rightarrow - \frac{1}{{{{\rm{R}}_{\rm{e}}} + {\rm{h}}}} + \frac{1}{{{{\rm{R}}_{\rm{e}}}}} = \frac{1}{{2\left( {{{\rm{R}}_{\rm{e}}} + {\rm{h}}} \right)}}\)

\(\Rightarrow \frac{1}{{{{\rm{R}}_{\rm{e}}}}} = \frac{1}{{2\left( {{{\rm{R}}_{\rm{e}}} + {\rm{h}}} \right)}} + \frac{1}{{\left( {{{\rm{R}}_{\rm{e}}} + {\rm{h}}} \right)}}\)

\(\Rightarrow \frac{1}{{{{\rm{R}}_{\rm{e}}}}} = \frac{{1 + 2}}{{2\left( {{{\rm{R}}_{\rm{e}}} + {\rm{h}}} \right)}}\)

⇒ 2R_e + h = 3R_e

⇒ 2R_e + 2h = 3R_e

⇒ 2h = 3R_e - 2R_e

⇒ 2h = R_e

\(\Rightarrow {\rm{h}} = \frac{{{{\rm{R}}_{\rm{e}}}}}{2}\)

We know that, R_e = 6.4 × 10³ km

\(\Rightarrow {\rm{h}} = \frac{{6.4{\rm{\;}} \times {\rm{\;}}{{10}^3}}}{2}\)

∴ h = 3.2 × 10³ km