Work in a gravitational field For large distances from the...

Work in a gravitational field For large distances from the surface of Earth, the gravitational force is given by $F(x)=\frac{G M m}{(x+R)^{2}}$ where $G=6.7 \times 10^{-11} \mathrm{N} \mathrm{m}^{2} / \mathrm{kg}^{2}$ is the gravitational constant. $M=6 \times 10^{24} \mathrm{kg}$ is the mass of Earth, $m$ is the mass of the object in the gravitational field, $R=6.378 \times 10^{6} \mathrm{m}$ is the radius of Farth, and $x \geq 0$ is the distance above the surface of Earth (in meters). a. How much work is required to launch a rocket with a mass of $500 \mathrm{kg}$ in a vertical flight path to a height of 2500 km (from Earth's surface)? b. Find the work required to launch the rocket to a height of $x$ kilometers, for $x>0$ c. How much work is required to reach outer space $(x \rightarrow \infty) ?$ d. Equate the work in part (c) to the initial kinetic energy of the rocket. $\frac{1}{2} m v^{2},$ to compute the escape velocity of the rocket.

Work in a gravitational field For large distances from the surface of Earth, the gravitational force is given by $F(x)=\frac{G M m}{(x+R)^{2}}$ where $G=6.7 \times 10^{-11} \mathrm{N} \mathrm{m}^{2} / \mathrm{kg}^{2}$ is the gravitational constant. $M=6 \times 10^{24} \mathrm{kg}$ is the mass of Earth, $m$ is the mass of the object in the gravitational field, $R=6.378 \times 10^{6} \mathrm{m}$ is the radius of Farth, and $x \geq 0$ is the distance above the surface of Earth (in meters).
a. How much work is required to launch a rocket with a mass of $500 \mathrm{kg}$ in a vertical flight path to a height of 2500 km (from Earth's surface)?
b. Find the work required to launch the rocket to a height of $x$ kilometers, for $x>0$
c. How much work is required to reach outer space $(x \rightarrow \infty) ?$
d. Equate the work in part (c) to the initial kinetic energy of the rocket. $\frac{1}{2} m v^{2},$ to compute the escape velocity of the rocket.

Key Concepts

Conservation of Energy

The conservation of energy principle states that the total energy in a closed system remains constant over time. In gravitational systems, this means that the sum of kinetic energy and gravitational potential energy is conserved. This principle is used to relate the work done or energy required to move a mass in a gravitational field to the velocity necessary to achieve such a displacement, particularly in deriving escape velocities.

Escape Velocity

Escape velocity is the minimum speed required for an object to break free from the gravitational attraction of a massive body without further propulsion. It is derived by equating the kinetic energy needed with the gravitational potential energy at an infinite distance. This concept is critical in astrophysics and space exploration, as it determines the energy thresholds for launching spacecraft beyond the influence of a planet's gravity.

Inverse-Square Law

The inverse-square law is a fundamental principle describing the manner in which a physical quantity or strength decreases with the square of the distance from the source. In the context of gravitation, it means that the gravitational force diminishes as the square of the distance increases, necessitating an integration approach when calculating work over large distances.

Work Done Against Gravity

Work in a gravitational field is defined as the energy required to move an object against the gravitational force. It is calculated by integrating the force over the distance moved along the field. In varying gravitational fields, such as those described by an inverse-square law, the integration involves more complex expressions than the constant force approximation, reflecting the changes in force with position.

Gravitational Force

This is the attractive force between two masses described by Newton’s law of universal gravitation. It states that the force between two objects is proportional to the product of their masses and inversely proportional to the square of the distance between their centers. The concept underpins many calculations in celestial mechanics and is essential when considering how objects move in and escape from gravitational fields.

Gravitational Potential Energy

Gravitational potential energy is the energy stored in an object due to its position in a gravitational field. It represents the work done to move the object from a reference point (often taken as infinity or the Earth’s surface) to a certain distance from the mass responsible for the field. This concept is crucial for solving problems involving lifting objects to high altitudes or determining the energy required for an object to escape the gravitational influence.

Transcript

00:03 In this question, we are given the force of gravity formula with a few constants and values.

00:10 We are asked to find the work required to launch a 500 -kilogram rocket up by 2 ,500 kilometers.

00:18 So first, we do want the distance to be in meters.

00:22 So i will just rewrite this as 2 .5 times 10 to the 6 meters, because i'm adding 3 zeros in order to convert it to meters.

00:31 And next we want to write the work formula, which is work equals integral from a to b, f of x, dx.

00:42 And we have this f function with a lot of letters, but the only variable is x.

00:53 So this is from 0 to 2 .5 times 10 to 6, gm over x plus r squared d.

01:06 And since gm and m are all constants, we can put those outside of the integral.

01:14 So it will be g capital m small m integral from 0 to 2 .5 times 10 to the 6.

01:22 And instead of writing 1 over x plus r squared, i can write this as x plus r to the power of negative 2.

01:33 So now i can just use the power rule for the integral.

01:35 So g m m there'll be x plus r to the negative one divided by negative one.

01:47 And i'm evaluating from 0 to 2 .5 times 10 to the 6.

01:53 The negative 1, i can, this denominator negative 1, i can just move it to the front.

01:59 And this exponent negative 1, i can just change this back to 1 over x plus r.

02:05 So this is negative g, big m, small m, 1, 1 over.

02:10 X plus r and i'm evaluating from 0 to 2 .5 times 10 to the 6 which is just this and then times 1 over 2 .5 times 10 to the 6 plus r minus 1 over and then when i put in 0 of into x my second term would just be 1 over r and if i were to put in all the values that are given the g capital m, 500 for small m and the r, i would get a .87 times 10 to the 9 joules.

02:59 Second question asks us to find the work to launch the 500 kg rocket up by x kilometer without specifying what x is.

03:10 So we have basically done that already in this formula, sorry, in this calculation.

03:18 The only difference is that we're not using x.

03:21 We're not using 2 .5 times 10 to 6.

03:24 We're using x.

03:27 So we just rewrite that here, which is negative gmm, 1 over x plus r.

03:38 And we are evaluating this from 0 to x instead.

03:43 So this would be negative gmm 1 over.

03:50 So x replacing x, which is still just being an x.

03:55 And 0 putting in there would be, again, just 1 over r.

04:01 And that's basically all you need to do...