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PhyHardAdv 2017 • Paper 2

Minimum Initial Escape Velocity to Leave Sun Earth System

Suppose a rocket is launched perpendicular to the surface of the Earth, directly away from the Sun, along the axis joining the Sun and the Earth. The mass of the Sun is 3×1053 \times 10^5 times that of the Earth, and its distance is 2.5×1042.5 \times 10^4 times larger than the radius of the Earth. Given that the escape speed from Earth's gravitational field is ve=11.2 km s−1v_e = 11.2\text{ km s}^{-1}, the minimum initial velocity (vSv_S) required for the rocket to completely leave the Sun-Earth system is nearest to: (Disregard the orbital and rotational motion of the Earth and the gravitational influence of other planets)

Options

A

vS=22 km s−1v_S = 22\text{ km s}^{-1}

B

vS=42 km s−1v_S = 42\text{ km s}^{-1}

Correct
C

vS=62 km s−1v_S = 62\text{ km s}^{-1}

D

vS=72 km s−1v_S = 72\text{ km s}^{-1}

Topics & Concepts

GravitationGravitation

Step-by-Step Solution

To find the minimum initial velocity vSv_S required for the rocket to completely leave the Sun-Earth system, we apply the law of conservation of mechanical energy.

1. Potential Energy at the Launch Point

Let:

  • MEM_E be the mass of the Earth,
  • RER_E be the radius of the Earth,
  • MS=3×105MEM_S = 3 \times 10^5 M_E be the mass of the Sun,
  • d=2.5×104REd = 2.5 \times 10^4 R_E be the distance between the Sun and the Earth.

The rocket is launched from the surface of the Earth away from the Sun, along the line joining the Sun and the Earth. Therefore, the distance of the rocket from:

  • The center of the Earth is rE=REr_E = R_E
  • The center of the Sun is rS=d+REr_S = d + R_E

The total gravitational potential energy per unit mass at the surface of the Earth is: Vi=−GMERE−GMSd+REV_i = -\frac{G M_E}{R_E} - \frac{G M_S}{d + R_E}

Substituting MS=3×105MEM_S = 3 \times 10^5 M_E and d=2.5×104REd = 2.5 \times 10^4 R_E: Vi=−GMERE(1+3×1052.5×104+1)V_i = -\frac{G M_E}{R_E} \left( 1 + \frac{3 \times 10^5}{2.5 \times 10^4 + 1} \right)

Since 2.5×104≫12.5 \times 10^4 \gg 1, we can approximate 2.5×104+1≈2.5×1042.5 \times 10^4 + 1 \approx 2.5 \times 10^4: 3×1052.5×104=12\frac{3 \times 10^5}{2.5 \times 10^4} = 12

Thus, the potential energy per unit mass simplifies to: Vi≈−GMERE(1+12)=−13GMEREV_i \approx -\frac{G M_E}{R_E} (1 + 12) = -13 \frac{G M_E}{R_E}

2. Conservation of Mechanical Energy

To leave the Sun-Earth system completely, the rocket must reach infinity (r→∞r \to \infty) where the gravitational potential energy V∞=0V_\infty = 0. For minimum initial velocity vSv_S, the kinetic energy at infinity is zero (K∞=0K_\infty = 0).

By conservation of energy: Ki+Vi=K∞+V∞K_i + V_i = K_\infty + V_\infty 12vS2−13GMERE=0\frac{1}{2} v_S^2 - 13 \frac{G M_E}{R_E} = 0 12vS2=13GMERE\frac{1}{2} v_S^2 = 13 \frac{G M_E}{R_E}

3. Calculation of vSv_S

We know that the escape velocity from the surface of Earth alone is given by: ve=2GMERE=11.2 km s−1  ⟹  GMERE=ve22v_e = \sqrt{\frac{2 G M_E}{R_E}} = 11.2\text{ km s}^{-1} \implies \frac{G M_E}{R_E} = \frac{v_e^2}{2}

Substituting this into the energy equation: 12vS2=13(ve22)\frac{1}{2} v_S^2 = 13 \left( \frac{v_e^2}{2} \right) vS2=13ve2v_S^2 = 13 v_e^2 vS=13vev_S = \sqrt{13} v_e

Using 13≈3.606\sqrt{13} \approx 3.606: vS≈3.606×11.2 km s−1≈40.38 km s−1v_S \approx 3.606 \times 11.2\text{ km s}^{-1} \approx 40.38\text{ km s}^{-1}

Among the given choices, 40.38 km s−140.38\text{ km s}^{-1} is closest to 42 km s−142\text{ km s}^{-1}.

Correct Option: [B] vS=42 km s−1v_S = 42\text{ km s}^{-1}

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