The mass of a spaceship is 1000 kg. It is to be launched from the earth's surface out into free spac

Q: Why is the potential energy at infinity considered to be zero?

It's a standard convention in physics. As the distance between two masses becomes infinitely large, the gravitational force between them approaches zero, making it a convenient reference point for zero potential energy.

Q: Why is the potential energy on the Earth's surface negative?

Gravitational force is always attractive. Because we define potential energy as zero at infinity, any point closer than infinity must have a lower (negative) potential energy. It means you have to do positive work against gravity to separate the masses.

Q: Can I just use the formula $mgh$ to find the energy?

No, the formula $U = mgh$ is an approximation that only works for small heights near the Earth's surface where $g$ is considered constant. For launching into free space, $g$ decreases significantly, so we must use the general formula $U = -GMm/r$.

Q: How does this required energy relate to escape velocity?

The required energy we calculated is exactly the initial kinetic energy ($\frac{1}{2}mv_e^2$) the spaceship needs to escape Earth's gravity. If you equate them, you can solve for the escape velocity $v_e$.

Q: Why did we substitute $GM$ with $gR^2$?

The problem didn't provide the values for $G$ (universal gravitational constant) or $M$ (mass of the Earth). However, it did provide $g$ and $R$. Using the relation $g = GM/R^2$ allows us to express the energy purely in terms of the given variables.

Question

The mass of a spaceship is $1000 	ext{ kg}$. It is to be launched from the earth's surface out into free space. The value of $g$ and $R$ (radius of earth) are $10 	ext{ m/s}^2$ and $6400 	ext{ km}$, respectively. The required energy for this work will be

Options:

(A) $6.4 	imes 10^{11} 	ext{ J}$

(B) $6.4 	imes 10^8 	ext{ J}$

(C) $6.4 	imes 10^9 	ext{ J}$

(D) $6.4 	imes 10^{10} 	ext{ J}$

Sigma Percentile · Accepted Answer

### The Mission Brief

Imagine you are the flight director at a space agency. Your mission? To launch a $1000 	ext{ kg}$ spaceship from the surface of the Earth out into the vast emptiness of free space. You are given the acceleration due to gravity, $g = 10 	ext{ m/s}^2$, and the radius of the Earth, $R = 6400 	ext{ km}$. Your task is to calculate the exact amount of energy required to achieve this monumental feat.

### The Physics of Escape

To break free from Earth's gravitational grip, we need to overcome its potential energy well. The energy required to send the spaceship to infinity (free space) is simply the difference between its final potential energy and its initial potential energy.

$$E_{	ext{required}} = U_{	ext{final}} - U_{	ext{initial}}$$

By convention, the gravitational potential energy at an infinite distance is zero ($U_{	ext{final}} = 0$). On the surface of the Earth, the potential energy is given by the formula:

$$U_{	ext{initial}} = -\frac{GMm}{R}$$

Here, $G$ is the universal gravitational constant, $M$ is the mass of the Earth, and $m$ is the mass of the spaceship.

### The Gravity Trick

Now, you might be thinking, 'We don't know the mass of the Earth or the gravitational constant!' Don't worry, we have a clever trick up our sleeve. We know the relationship between the acceleration due to gravity $g$ and these constants:

$$g = \frac{GM}{R^2}$$

Rearranging this, we get $GM = gR^2$. Let's substitute this back into our initial potential energy equation:

$$U_{	ext{initial}} = -\frac{(gR^2)m}{R} = -mgR$$

This beautifully simplifies our required energy equation:

$$E_{	ext{required}} = 0 - (-mgR) = mgR$$

### Crunching the Numbers

We are now ready to plug in the numbers. But first, a crucial step: **unit conversion**. The radius of the Earth is given in kilometers, so we must convert it to meters to match the SI units of $g$.

$$R = 6400 	ext{ km} = 6.4 	imes 10^6 	ext{ m}$$

Now, let's substitute everything into our simplified energy equation:

$$E_{	ext{required}} = (1000 	ext{ kg}) 	imes (10 	ext{ m/s}^2) 	imes (6.4 	imes 10^6 	ext{ m})$$

### Mission Accomplished

Multiplying these values together, we get:

$$E_{	ext{required}} = 6.4 	imes 10^{10} 	ext{ J}$$

And there we have it! The required energy to launch the spaceship into free space is **$6.4 	imes 10^{10} 	ext{ J}$**. This energy is exactly what would be provided as kinetic energy if we were calculating the escape velocity of the spaceship. Physics is truly interconnected and elegant!

Animated Solution for Physics - Gravitation: The mass of a spaceship is $1000 kg$ . It is to be launched from the earth's surface out into free space. The value of $g$ and $R$ (radius of earth) are $10 m/s^{2}$ and $6400 km$ , respectively. The required energy for this work will be

Select Answer:

Visualized Solution

$Given Data$

$Energy Required$

$Simplifying Potential Energy$

$Substitution$

$Final Calculation$

$Concept Extension$

The Sigma Insight: Gravitational Potential and Potential Energy

The Mission Brief

The Physics of Escape

The Gravity Trick

Crunching the Numbers

Mission Accomplished

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Given Data

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

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Substitution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Calculation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Concept Extension