A rocket is launched normal to the surface of the Earth, away from the Sun, along the line joining t

Q: Why can we approximate the distance of the rocket from the Sun as $r$ instead of $r + R_e$?

Since the distance between the Sun and the Earth ($r = 2.5 \times 10^4 R_e$) is extremely large compared to the Earth's radius ($R_e$), the difference between $r$ and $r + R_e$ is negligible (less than $0.004\%$). Thus, approximating the distance as $r$ is highly accurate.

Q: Why is the potential energy due to both the Earth and the Sun negative at the launch point?

Gravitational potential energy is always negative because gravity is an attractive force, and we define the potential energy to be zero at an infinite separation. Work must be done against gravity to move the rocket away, which increases its potential energy towards zero.

Q: Does the rocket's mass affect the required escape velocity?

No. In the energy conservation equation, the mass of the rocket $m$ appears in every term (both kinetic and potential energy). Dividing the entire equation by $m$ eliminates it, showing that escape velocity depends only on the masses and positions of the celestial bodies.

Q: Why did we ignore the orbital motion of the Earth around the Sun?

The problem explicitly instructs us to 'Ignore the rotation and revolution of the Earth'. In reality, launching in the direction of Earth's orbital motion would give the rocket an initial boost of about $30 \text{ km/s}$ relative to the Sun, significantly reducing the required launch speed from Earth's surface.

Q: How do we know the total mechanical energy at infinity is zero?

For the 'minimum' velocity required to escape, the rocket should just barely reach infinity, meaning its velocity (and thus kinetic energy) at infinity approaches zero. Since potential energy at infinity is also defined as zero, the total mechanical energy is $0 + 0 = 0$.

Question

A rocket is launched normal to the surface of the Earth, away from the Sun, along the line joining the Sun and the Earth. The Sun is $3 	imes 10^5$ times heavier than the Earth and is at a distance $2.5 	imes 10^4$ times larger than the radius of Earth. The escape velocity from Earth's gravitational field is $v_e = 11.2 	ext{ km s}^{-1}$. The minimum initial velocity ($v_s$) required for the rocket to be able to leave the Sun-Earth system is closest to (Ignore the rotation and revolution of the Earth and the presence of any other planet)

Options:

(A) $v_s = 72 	ext{ km s}^{-1}$

(B) $v_s = 22 	ext{ km s}^{-1}$

(C) $v_s = 42 	ext{ km s}^{-1}$

(D) $v_s = 62 	ext{ km s}^{-1}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine standing on the surface of the Earth, looking up at the night sky, away from the Sun.

We are launching a rocket vertically upwards—directly along the line connecting the Sun and the Earth, but heading outward into deep space.

To escape the gravitational pull of our cosmic neighborhood, the rocket must overcome not just the Earth's gravity, but also the immense gravitational field of the Sun.

Let's define our parameters clearly:

- Mass of the Earth: $M_e$
- Radius of the Earth: $R_e$
- Mass of the Sun: $M_s = 3 	imes 10^5 M_e$
- Distance between the Sun and the Earth: $r = 2.5 	imes 10^4 R_e$

### The Master Equation

To find the minimum launch velocity $v_s$ required to escape to infinity, we apply the **Conservation of Mechanical Energy**.

At infinity, both the kinetic energy and the gravitational potential energy of the rocket relative to both bodies will be zero.

Therefore, the total mechanical energy at infinity is:

$$E_{\infty} = 0$$

By conservation of energy, the total mechanical energy at the launch point must also be zero:

$$K_i + U_i = 0$$

Let's write down the individual energy terms at the surface of the Earth:

- Kinetic Energy: $\frac{1}{2} m v_s^2$
- Gravitational Potential Energy due to Earth: $-\frac{G M_e m}{R_e}$
- Gravitational Potential Energy due to the Sun: $-\frac{G M_s m}{r}$

Combining these into our energy conservation equation:

$$\frac{1}{2} m v_s^2 - \frac{G M_e m}{R_e} - \frac{G M_s m}{r} = 0$$

### Solving for Escape Velocity

We can divide out the mass of the rocket $m$, showing that escape velocity is independent of the mass of the projectile itself:

$$\frac{1}{2} v_s^2 = \frac{G M_e}{R_e} + \frac{G M_s}{r}$$

Multiplying by $2$ and taking the square root gives us the expression for the escape velocity:

$$v_s = \sqrt{\frac{2GM_e}{R_e} + \frac{2GM_s}{r}}$$

Notice that the first term under the square root, $\frac{2GM_e}{R_e}$, is exactly the square of the Earth's standard escape velocity $v_e^2$:

$$v_e = \sqrt{\frac{2GM_e}{R_e}} = 11.2 	ext{ km/s}$$

Let's factor this term out to make our calculation elegant:

$$v_s = \sqrt{\frac{2GM_e}{R_e} \left[ 1 + \frac{M_s}{M_e} \cdot \frac{R_e}{r} ight]}$$

### Substituting the Values

Now, let's plug in the given ratios for the Sun's mass and distance:

$$\frac{M_s}{M_e} = 3 	imes 10^5$$

$$\frac{r}{R_e} = 2.5 	imes 10^4 \implies \frac{R_e}{r} = \frac{1}{2.5 	imes 10^4}$$

Substituting these ratios into the bracket:

$$	ext{Ratio} = \frac{3 	imes 10^5}{2.5 	imes 10^4} = \frac{30}{2.5} = 12$$

This simplifies our escape velocity equation beautifully:

$$v_s = v_e \sqrt{1 + 12} = v_e \sqrt{13}$$

### Final Calculation

Using the given value of Earth's escape velocity $v_e = 11.2 	ext{ km/s}$:

$$v_s = 11.2 	imes \sqrt{13} 	ext{ km/s}$$

Since $\sqrt{13} \approx 3.6055$:

$$v_s \approx 11.2 	imes 3.6055 \approx 40.38 	ext{ km/s}$$

Comparing this with the given options:

- (a) $72 	ext{ km/s}$
- (b) $22 	ext{ km/s}$
- (c) $42 	ext{ km/s}$
- (d) $62 	ext{ km/s}$

The calculated value of $40.38 	ext{ km/s}$ is closest to **$42 	ext{ km/s}$**.

Thus, the correct option is **(c)**.

Select Answer:

Visualized Solution

Visualizing the Sun-Earth-Rocket System

Defining Earth's Escape Velocity

Applying Conservation of Mechanical Energy

Setting Up the Energy Equation

Solving for Launch Velocity $v_{s}$

Substituting Given Ratios

Factoring Out Earth's Escape Velocity

Calculating the Final Value

Matching with the Closest Option

The Sigma Insight: Gravitational Potential and Potential Energy

Analyzing the Setup

The Master Equation

Solving for Escape Velocity

Substituting the Values

Final Calculation

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