Two stars of masses 3 x 10^31 kg each and at distance 2 x 10^11 m rotate in a plane about their comm

Q: Why do we set the total energy to zero for escape velocity?

When an object escapes a gravitational field, it reaches infinity. At infinity, the gravitational potential energy is zero. The *minimum* speed required to just reach infinity implies it arrives there with zero kinetic energy. Therefore, the total mechanical energy (KE + PE) must be exactly zero.

Q: Does the direction of the meteorite matter for escape velocity?

Interestingly, no! Escape velocity is derived from energy conservation, which involves scalar quantities (Kinetic and Potential Energy). As long as the meteorite doesn't crash into one of the stars, any direction will allow it to escape if it has the required speed.

Q: Why did we use $R$ instead of $2R$ in the potential energy formula?

The formula for potential energy between two masses is $-GMm/r$, where $r$ is the distance *between* them. The problem states the distance between the two stars is $2 \times 10^{11}$ m. Since the meteorite is at the center of mass, it is exactly halfway between them, making its distance to *each* star $R = 10^{11}$ m.

Q: Why doesn't the mass of the meteorite affect the final answer?

Both the kinetic energy ($\frac{1}{2}mv^2$) and the potential energy ($-\frac{GMm}{r}$) are directly proportional to the mass of the meteorite, $m$. When you set their sum to zero, the mass $m$ cancels out of the equation entirely. This means a tiny pebble and a massive spaceship would need the exact same speed to escape from that specific point!

Q: What if the meteorite was not exactly at the center?

If the meteorite were closer to one star, its initial potential energy would be more negative (a deeper 'potential well'). Consequently, it would require a higher initial kinetic energy, and thus a higher escape speed, to break free from the system.

Question

Two stars of masses $3 	imes 10^{31} 	ext{ kg}$ each and at distance $2 	imes 10^{11} 	ext{ m}$ rotate in a plane about their common centre of mass $O$. A meteorite passes through $O$ moving perpendicular to the star's rotation plane. In order to escape from the gravitational field of this double star, the minimum speed that meteorite should have at $O$ is (Take, gravitational constant, $G = 6.67 	imes 10^{-11} 	ext{ N-m}^2	ext{kg}^{-2}$)

Options:

(A) $2.8 	imes 10^5 	ext{ m/s}$

(B) $3.8 	imes 10^4 	ext{ m/s}$

(C) $2.4 	imes 10^4 	ext{ m/s}$

(D) $1.4 	imes 10^5 	ext{ m/s}$

Sigma Percentile · Accepted Answer

### The Cosmic Dance and the Great Escape

Imagine you are standing at the exact center of a colossal cosmic dance. On either side of you, at equal distances, are two incredibly massive stars, each weighing a staggering $3 	imes 10^{31} 	ext{ kg}$. They are locked in a gravitational embrace, orbiting their common center of mass—the very point where you are standing.

Now, imagine a meteorite passing through this exact center point, moving perpendicular to the plane in which these stars are orbiting. The question we need to answer is: **How fast must this meteorite be traveling to completely escape the gravitational clutches of both stars and drift off into the infinite void?**

This is a classic problem of escape velocity, but with a fascinating twist—we are dealing with a binary star system instead of a single planet.

### Analyzing the Potential Well

To understand escape velocity, we must first understand the concept of a gravitational potential well. Any object with mass creates a 'dip' in the gravitational potential around it. To escape, an object must have enough kinetic energy to climb completely out of this well.

At the center of mass $O$, the meteorite is at the very bottom of a combined potential well created by *both* stars. Since gravitational potential is a scalar quantity, we can simply add the potential energies contributed by each star.

Let the mass of each star be $M$ and the distance from the center $O$ to each star be $R$. The total gravitational potential energy $U_{	ext{total}}$ of the meteorite (mass $m$) at point $O$ is:

$$U_{	ext{total}} = U_1 + U_2$$

$$U_{	ext{total}} = -\frac{GMm}{R} - \frac{GMm}{R} = -\frac{2GMm}{R}$$

This negative value represents how tightly the meteorite is bound to the system.

### The Master Equation: Conservation of Energy

To escape to infinity, the meteorite must reach a point where the gravitational pull is zero (meaning potential energy is zero). The *minimum* speed required to do this implies that the meteorite arrives at infinity with exactly zero kinetic energy left over.

Therefore, the total mechanical energy (Kinetic Energy + Potential Energy) of the meteorite must be exactly zero.

Let $v$ be the escape speed. The initial kinetic energy is $K = \frac{1}{2}mv^2$. Applying the conservation of energy:

$$K + U_{	ext{total}} = 0$$

$$\frac{1}{2}mv^2 - \frac{2GMm}{R} = 0$$

Notice a beautiful piece of physics here: the mass of the meteorite, $m$, appears in both terms and cancels out completely!

$$\frac{1}{2}v^2 = \frac{2GM}{R}$$

$$v = \sqrt{\frac{4GM}{R}}$$

This tells us that whether it's a tiny pebble or a massive spaceship, the speed required to escape from that specific point is exactly the same.

### Final Calculation: Crunching the Numbers

Now, we just need to carefully substitute the given values into our derived formula.

**A word of caution:** The problem states the distance *between* the stars is $2 	imes 10^{11} 	ext{ m}$. This is $2R$. Therefore, the distance from the center to each star is $R = 10^{11} 	ext{ m}$.

Given values:
*   $M = 3 	imes 10^{31} 	ext{ kg}$
*   $R = 10^{11} 	ext{ m}$
*   $G = 6.67 	imes 10^{-11} 	ext{ N-m}^2	ext{kg}^{-2}$

Substituting these into our equation:

$$v = \sqrt{\frac{4 	imes 6.67 	imes 10^{-11} 	imes 3 	imes 10^{31}}{10^{11}}}$$

Let's handle the powers of 10 first: $10^{-11} 	imes 10^{31} = 10^{20}$. Dividing by $10^{11}$ leaves $10^9$.

$$v = \sqrt{4 	imes 6.67 	imes 3 	imes 10^9}$$

$$v = \sqrt{80.04 	imes 10^9}$$

To make taking the square root easier, let's adjust the decimal to get an even power of 10:

$$v = \sqrt{8.004 	imes 10^{10}}$$

Since $\sqrt{8} \approx 2.828$, we have:

$$v \approx 2.83 	imes 10^5 	ext{ m/s}$$

Rounding to one decimal place, we get **$2.8 	imes 10^5 	ext{ m/s}$**, which perfectly matches option (a). The meteorite must travel at this incredible speed to break free from the cosmic dance of the twin stars!

Select Answer:

Visualized Solution

Visualizing the Binary Star System

Gravitational Potential Energy at $O$

Condition for Escape

Expression for Escape Speed

Substituting the Values

Final Calculation

Conclusion

The Sigma Insight: Gravitational Potential and Potential Energy

The Cosmic Dance and the Great Escape

Analyzing the Potential Well

The Master Equation

Final Calculation

Similar Questions

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Two bodies, each of mass $M$ , are kept fixed with a separation $2 L$ . A particle of mass $m$ is projected from the mid-point of the line joining their centres, perpendicular to the line. The gravitational constant is $G$ . The correct statement(s) is (are)

A planet in a distant solar system is 10 times more massive than the earth and its radius is 10 times smaller. Given that the escape velocity from the earth is $11 kms^{- 1}$ , the escape velocity from the surface of the planet would be

Energy required to move a body of mass $m$ from an orbit of radius $2 R$ to $3 R$ is

Select Answer:

Visualized Solution

Visualizing the Binary Star System

Gravitational Potential Energy at .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } O

Condition for Escape

Expression for Escape Speed

Substituting the Values

Final Calculation

Conclusion

The Sigma Insight: Gravitational Potential and Potential Energy

The Cosmic Dance and the Great Escape

Analyzing the Potential Well

The Master Equation

Final Calculation

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Energy required to move a body of mass m from an orbit of radius 2R to 3R is

Gravitational Potential Energy at $O$

Energy required to move a body of mass $m$ from an orbit of radius $2 R$ to $3 R$ is