Solution Found!
The equation mgy for gravitational potential energy is
Chapter 11, Problem 74CP(choose chapter or problem)
The equation \(m g y\) for gravitational potential energy is valid only for objects near the surface of a planet. Consider two very large objects of mass \(m_{1}\) and \(m_{2}\), such as stars or planets, whose centers are separated by the large distance \(r\). These two large objects exert gravitational forces on each other. You'll learn in Chapter 13 that the gravitational potential energy is
\(U=-\frac{G m_{1} m_{2}}{r}\)
where \(G=6.67 \times 10^{-11} \mathrm{~N} \mathrm{~m}^{2} / \mathrm{kg}^{2}\) is the gravitational constant.
a. Sketch a graph of \(U\) versus \(r\). The mathematical difficulty at \(r=0\) is not a physically significant difficulty because the masses will collide before they get that close together.
b. What separation \(r\) has been chosen as the point of zero potential energy? Does this make sense? Explain.
c. Two stars are at rest \(1.0 \times 10^{14} \mathrm{~m}\) apart. This is about 10 times the diameter of the solar system. The first star is the size of our sun, with a mass of \(2.0 \times 10^{30} \mathrm{~kg}\) and a radius of \(7.0 \times 10^{8} \mathrm{~m}\). The second star has mass \(8.0 \times 10^{30} \mathrm{~kg}\) and radius of \(11.0 \times 10^{8} \mathrm{~m}\). Gravitational forces pull the two stars together. What is the speed of each star at the moment of impact?
Equation Transcription:
Text Transcription:
mgy
m_{1}
m_{2}
r
U = -frac{G m_{1} m_{2}}{r}
G = 6.67 X 10^{-11} Nm^{2} / kg^2
U
r
r = 0
r
1.0 X 10^{14} m
2.0 X 10^{30} kg
7.0 X 10^{8} m
8.0 Xs 10^{30} kg
11.0 X 10^{8} m
Questions & Answers
QUESTION:
The equation \(m g y\) for gravitational potential energy is valid only for objects near the surface of a planet. Consider two very large objects of mass \(m_{1}\) and \(m_{2}\), such as stars or planets, whose centers are separated by the large distance \(r\). These two large objects exert gravitational forces on each other. You'll learn in Chapter 13 that the gravitational potential energy is
\(U=-\frac{G m_{1} m_{2}}{r}\)
where \(G=6.67 \times 10^{-11} \mathrm{~N} \mathrm{~m}^{2} / \mathrm{kg}^{2}\) is the gravitational constant.
a. Sketch a graph of \(U\) versus \(r\). The mathematical difficulty at \(r=0\) is not a physically significant difficulty because the masses will collide before they get that close together.
b. What separation \(r\) has been chosen as the point of zero potential energy? Does this make sense? Explain.
c. Two stars are at rest \(1.0 \times 10^{14} \mathrm{~m}\) apart. This is about 10 times the diameter of the solar system. The first star is the size of our sun, with a mass of \(2.0 \times 10^{30} \mathrm{~kg}\) and a radius of \(7.0 \times 10^{8} \mathrm{~m}\). The second star has mass \(8.0 \times 10^{30} \mathrm{~kg}\) and radius of \(11.0 \times 10^{8} \mathrm{~m}\). Gravitational forces pull the two stars together. What is the speed of each star at the moment of impact?
Equation Transcription:
Text Transcription:
mgy
m_{1}
m_{2}
r
U = -frac{G m_{1} m_{2}}{r}
G = 6.67 X 10^{-11} Nm^{2} / kg^2
U
r
r = 0
r
1.0 X 10^{14} m
2.0 X 10^{30} kg
7.0 X 10^{8} m
8.0 Xs 10^{30} kg
11.0 X 10^{8} m
ANSWER:
Step 1 of 3
a) We need to sketch a U versus r graph.