1 mpe \( =1 / 2 \) (s) \( 2 \quad \operatorname{Re}=\frac{m p e}{m v} \) \( 3 \quad P P e=\operatorname{Re} x 100 \) EXAMPLE: \[ \frac{(5.628 \times 7-96 \times 4.4) \times 10^{2+1-4}}{(9.78 \times 8.2 \times 1.8) \times 10^{1+1-3}} \]
Added by Favour C.
Close
Step 1
Identify the problem or task at hand. Show more…
Show all steps
Your feedback will help us improve your experience
Eduard Sanchez and 92 other Physics 101 Mechanics educators are ready to help you.
Ask a new question
Labs
Want to see this concept in action?
Explore this concept interactively to see how it behaves as you change inputs.
Key Concepts
Recommended Videos
\begin{equation} \begin{array}{l}{\text { (a) Newton's Law of Gravitation states that two bodies }} \\ {\text { with masses } m_{1} \text { and } m_{2} \text { attract each other with a force }}\end{array} \end{equation} \begin{equation} F=G \frac{m_{1} m_{2}}{r^{2}} \end{equation} \begin{equation} \begin{array}{l}{\text { where } r \text { is the distance between the bodies and } G} \\ {\text { is the gravitational constant. If one of the bodies is }} \\ {\text { fixed, find the work needed to move the other from }} \\ {r=a \text { to } r=b \text { . }}\end{array} \end{equation} \begin{equation} \begin{array}{l}{\text { (b) Compute the work required to launch a } 1000-\mathrm{kg}} \\ {\text { satellite vertically to a height of } 1000 \mathrm{km} . \text { You may }} \\ {\text { assume that the earth's mass is } 5.98 \times 10^{24} \mathrm{kg}}\\{\text { and is concentrated at its center. Take the }} \\ {\text { radius of the earth to be } 6.37 \times 10^{6} \mathrm{m} \text { and }} \\ {G=6.67 \times 10^{-11} \mathrm{N} \cdot \mathrm{m}^{2} / \mathrm{kg}^{2}}.\end{array} \end{equation}
Applications of Integration
Work
Given that, $\begin{aligned} r_{M} &=5.8 \times 10^{10} \mathrm{~m} \\ M_{s} &=1.99 \times 10^{30} \mathrm{~kg} \\ T_{M} &=? \\ T_{M}^{2} &=\left(\frac{4 \pi^{2}}{G M_{s}}\right) r^{-3} M \\ T_{M}^{2}=&\left[\frac{39.43}{\left(6.673 \times 10^{-11}\right)\left(1.99 \times 10^{30}\right)}\right]\left(5.810 \times 10^{10}\right)^{3} \\ & T_{M}^{2}=\left[\frac{39.43}{6.673 \times 1.99 \times 10^{-11}}\right](5.810) \\ & T_{M}^{2}=\frac{229.1371}{13.27} \times 10^{11} \\ T_{M}^{2} &=172.673 \times 10^{11} \\ & T_{M}=76.35 \times 10^{5} \mathrm{~s} \end{aligned}$
GRAVITATION
Gravitation
$\omega_{1}=\frac{\gamma M}{R^{2}}=\frac{6.67 \times 10^{-11} \times 5 \cdot 96 \times 10^{24}}{\left(6.37 \times 10^{6}\right)^{2}}=9.8 \mathrm{~m} / \mathrm{s}^{2}$ $\omega_{2}=\omega^{2} R=\left(\frac{2 \pi}{T}\right)^{2} R=\left(\frac{2 \times 22}{24 \times 3600 \times 7}\right)^{2} 6.37 \times 10^{6}=0.034 \mathrm{~m} / \mathrm{s}^{2}$ and $\omega_{3}=\frac{\gamma M_{S}}{R_{\text {mean }}^{2}}=\frac{6.67 \times 10^{-11} \times 1.97 \times 10^{30}}{\left(149.50 \times 10^{6} \times 10^{3}\right)^{2}}=5.9 \times 10^{-3} \mathrm{~m} / \mathrm{s}^{2}$ Then $\quad \omega_{1}: \omega_{2}: \omega_{3}=1: 0.0034: 0.0006$
Physical Fundamentals Of Mdchanics
Universal Gravitation
Recommended Textbooks
University Physics with Modern Physics
Physics: Principles with Applications
Fundamentals of Physics
Watch the video solution with this free unlock.
EMAIL
PASSWORD