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40P

Expert-verifiedFound in: Page 1

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?**

- 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 $1.33$ .
- 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 $2.00$ .
- The least initial mechanical energy required at launch if the projectile is to escape Earth is 0.

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.

**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.

** **

Now,

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

As

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

${K}_{i}=\frac{1}{2}m{v}^{2}\&\u200a\u200a{K}_{f}=0\phantom{\rule{0ex}{0ex}}\frac{1}{2}m{v}^{2}-\frac{GMm}{{R}_{E}}=0-\frac{GMm}{R}\phantom{\rule{0ex}{0ex}}$

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" $1.33{R}_{E}$ .

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.

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.

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