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### Fundamentals Of Physics

Book edition 10th Edition
Author(s) David Halliday
Pages 1328 pages
ISBN 9781118230718

# A projectile is shot directly away from Earth’s surface.Neglect the rotation of Earth. What multiple of Earth’s radius gives the radial distance a projectile reaches if (a) its initial speed is ${\mathbf{0}}{\mathbf{.}}{\mathbf{500}}$ of the escape speed from Earth and(b) its initial kinetic energy is ${\mathbf{0}}{\mathbf{.}}{\mathbf{500}}$ of the kinetic energy required to escape Earth?(c)What is the least initial mechanical energy required at launch if the projectile is to escape Earth?

1. The multiple of Earth’s radius ${\text{R}}_{\text{E}}$ gives the radial distance a projectile reaches if its initial speed is of the escape speed from Earth is $1.33$ .
2. The multiple of Earth’s radius ${R}_{E}$ gives the radial distance a projectile reaches if its initial kinetic energy is $0.500$ of the kinetic energy required to escape Earth is $2.00$ .
3. The least initial mechanical energy required at launch if the projectile is to escape Earth is 0.
See the step by step solution

## Step 1: Given

The initial speed of the projectile is $0.500$of the escape speeds from Earth.

The initial kinetic energy of the projectile is $0.500$ of the kinetic energy required to escape Earth.

## Step 2: Determining the concept

Using the principle of energy conservation, find the multiple of Earth’s radius which gives the radial distance a projectile reaches if its initial speed is of the escape speed from Earth and if its initial kinetic energy is of the kinetic energy required to escape Earth.According to the law of conservation of energy, energy can neither be created nor be destroyed.

Formulae are as follows:

${K}_{i}+{U}_{i}={K}_{f}+{U}_{f}\phantom{\rule{0ex}{0ex}}U=-\frac{GMm}{R}\phantom{\rule{0ex}{0ex}}K=\frac{1}{2}m{v}^{2}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

where, M, and m are masses, R is the radius, v is velocity, G is gravitational constant, K is kinetic energy and U is potential energy.

## Step 3: (a) Determining the multiple of earth’s radius  RE that gives the radial distance a projectile reaches if its initial speed is 0.500  of the escape speed from Earth

Now,

${K}_{i}+{U}_{i}={K}_{f}+{U}_{f}$

As

${U}_{i}=-\frac{GMm}{{R}_{E}}$and ${U}_{f}=-\frac{GMm}{R},$

${K}_{i}=\frac{1}{2}m{v}^{2}& {K}_{f}=0\phantom{\rule{0ex}{0ex}}\frac{1}{2}m{v}^{2}-\frac{GMm}{{R}_{E}}=0-\frac{GMm}{R}\phantom{\rule{0ex}{0ex}}$

As

${v}_{e}=\sqrt{\frac{2GM}{{R}_{E}}}\phantom{\rule{0ex}{0ex}}\frac{1}{8}m{\left(\sqrt{\frac{2GM}{{R}_{E}}}\right)}^{2}-\frac{GMm}{{R}_{E}}=0-\frac{GMm}{R}\phantom{\rule{0ex}{0ex}}\frac{2GMm}{8{R}_{E}}-\frac{GMm}{{R}_{E}}=-\frac{GMm}{R}\phantom{\rule{0ex}{0ex}}\frac{GMm}{4{R}_{E}}-\frac{GMm}{{R}_{E}}=-\frac{GMm}{R}\phantom{\rule{0ex}{0ex}}\frac{3}{4{R}_{E}}=\frac{1}{R}\phantom{\rule{0ex}{0ex}}\frac{{R}_{E}}{R}=\frac{3}{4}\phantom{\rule{0ex}{0ex}}\frac{R}{{R}_{E}}=\frac{4}{3}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

$R=1.33{R}_{E}$

Therefore, the multiple of Earth’s radius gives the radial distance a projectile reaches if its initial speed is of the escape speed from Earth is role="math" localid="1661194870557" $1.33{R}_{E}$ .

## Step 4: (b) Determining the multiple of earth’s radius  RE that gives the radial distance a projectile reaches if its initial kinetic energy is  0.500 of the kinetic energy required to escape earth

Now,

${K}_{i}+{U}_{i}={K}_{f}+{U}_{f}$

As

.${U}_{i}=-\frac{GMm}{{R}_{E}}\phantom{\rule{0ex}{0ex}}{U}_{f}=-\frac{GMm}{R}\phantom{\rule{0ex}{0ex}}{K}_{i}=\frac{1}{2}m{v}^{2}\phantom{\rule{0ex}{0ex}}{K}_{f}=0\phantom{\rule{0ex}{0ex}}\frac{1}{2}m{v}^{2}-\frac{GMm}{{R}_{E}}=0-\frac{GMm}{R}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

As

$\frac{1}{2}m{v}^{2}=\frac{\left(\frac{1}{2}m{v}_{e}^{2}\right)}{2}\phantom{\rule{0ex}{0ex}}\frac{\left(\frac{1}{2}m{v}_{e}^{2}\right)}{2}-\frac{GMm}{{R}_{E}}=0-\frac{GMm}{R}\phantom{\rule{0ex}{0ex}}\frac{m{v}_{e}^{2}}{4}-\frac{GMm}{{R}_{E}}=0-\frac{GMm}{R}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

As

${v}_{e}=\sqrt{\frac{2GM}{{R}_{E}}}\phantom{\rule{0ex}{0ex}}\frac{1}{4}m{\left(\sqrt{\frac{2GM}{{R}_{E}}}\right)}^{2}-\frac{GMm}{{R}_{E}}=0-\frac{GMm}{R}\phantom{\rule{0ex}{0ex}}\frac{2GMm}{4{R}_{E}}-\frac{GMm}{{R}_{E}}=-\frac{GMm}{R}\phantom{\rule{0ex}{0ex}}\frac{1}{2}\frac{GMm}{{R}_{E}}-\frac{GMm}{{R}_{E}}=-\frac{GMm}{R}\phantom{\rule{0ex}{0ex}}-\frac{1}{2}\frac{GMm}{{R}_{E}}=-\frac{GMm}{R}\phantom{\rule{0ex}{0ex}}\frac{R}{{R}_{E}}=2\phantom{\rule{0ex}{0ex}}R=2{R}_{E}\phantom{\rule{0ex}{0ex}}$

Therefore, the multiple of Earth’s radius ${R}_{E}$ that gives the radial distance a projectile reaches if its initial kinetic energy is $0.500$ of the kinetic energy required to escape Earth.

## Step 5: (c) Determining the least initial mechanical energy  required at launch if the projectile is to escape Earth

Now,

${K}_{i}+{U}_{i}={K}_{f}+{U}_{f}$

As

${U}_{i}=-\frac{GMm}{{R}_{E}}\phantom{\rule{0ex}{0ex}}{\text{U}}_{\text{f}}=0,\phantom{\rule{0ex}{0ex}}{K}_{i}=\frac{1}{2}m{v}^{2}\phantom{\rule{0ex}{0ex}}{K}_{f}=0\phantom{\rule{0ex}{0ex}}\frac{1}{2}m{v}^{2}-\frac{GMm}{{R}_{E}}=0-0\phantom{\rule{0ex}{0ex}}\frac{1}{2}m{v}_{e}^{2}-\frac{GMm}{{R}_{E}}=0\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

As

${v}_{e}=\sqrt{\frac{2GM}{{R}_{E}}}\phantom{\rule{0ex}{0ex}}\frac{m}{2}{\left(\sqrt{\frac{2GM}{{R}_{E}}}\right)}^{2}-\frac{GMm}{{R}_{E}}=0\phantom{\rule{0ex}{0ex}}\frac{GMm}{{R}_{E}}-\frac{GMm}{{R}_{E}}=0\phantom{\rule{0ex}{0ex}}0=0\phantom{\rule{0ex}{0ex}}$

Hence, the least initial mechanical energy required at launch if the projectile is to escape Earth is 0 .

Therefore, using the formula for gravitational potential energy and kinetic energy along with the law of conservation of energy, the required distance can be found.