Chapter 12: Basic Physics SEGravity
The Big Idea
Over 2500 years ago Aristotle proposed two laws of physics governing motion. One for ‘Earthly bodies’ (objects on Earth) that states objects naturally go in straight lines and one for ‘Heavenly bodies’ (objects in space) that states objects naturally go in circles. This idea held steady for 2,000 years, until Isaac Newton in a triumph of brilliance declared that there is one law of physics that governs motion and he unified “earthly” bodies and “heavenly” bodies with the The Universal Law of Gravitation. Newton used the universal law of gravity to prove that the moon is accelerating towards Earth just as a body on Earth is falling towards Earth. The acceleration of both the moon and the object on Earth is calculated using the same formula. This theory is so well regarded that it has been used to hypothesize the existence of black holes and dark matter, later verified by observation. The search to unify the laws of physics into one theory continues today and is often referred to as the yet undiscovered Grand Unified Theory (GUT).
Key Equation: The Universal Law of Gravity
 \begin{align*}F_G =\frac{Gm_1m_2}{r^2}\end{align*} ; the force of gravity between an object with mass \begin{align*}m_1\end{align*} and another object of mass \begin{align*}m_2\end{align*} and a distance between them of \begin{align*}r\end{align*}.
 \begin{align*}G = 6.67 \times 10^{11} \ Nm^2 / kg^2\end{align*} ; the universal constant of gravity
 \begin{align*}g = \frac{Gm}{r^2}\end{align*} ; gravitational field strength or gravitational acceleration of a planet with mass \begin{align*}m\end{align*} and radius \begin{align*}r\end{align*}. Note that this is not really a separate equation but comes from Newton’s second law and the law of universal gravitation.
Key Concepts
 When using the Universal Law of Gravity formula and the constant \begin{align*}G\end{align*} above, make sure to use units of meters and kilograms.
 Newton invented calculus in order to prove that for a spherical object (like Earth) one can assume all of its mass is at the center of the sphere (thus in his formula, one can use the radius of Earth for the distance between a falling rock and Earth).
 An orbital period, \begin{align*}T\end{align*}, is the time it takes to make one complete rotation.
 If a particle travels a distance \begin{align*}2 \pi r\end{align*} in an amount of time \begin{align*}T\end{align*}, then its speed is distance over time or \begin{align*}\frac{2 \pi r}{T}\end{align*}.
 Objects in orbit around each other, orbit about the center of mass for the system. For the Earth and moon the center of mass is somewhere inside of Earth. So while the moon orbits Earth, the Earth also orbits the moon (manifested as a slight wobble).
Key Applications
 To find the speed of a planet or satellite in an orbit, realize that the force of gravity is the centripetal force. So set the force of gravity equal to \begin{align*}\frac{mv^2}{r}\end{align*}, where \begin{align*}v\end{align*} is the speed of the planet, \begin{align*}m\end{align*} is the mass of the planet or satellite and \begin{align*}r\end{align*} is the distance of the planet to the sun or the satellite to the Earth.
 The Geosynchronous orbit is that orbit for which the satellite takes the same amount of time to orbit the planet as the planet takes to make one revolution. For satellites to be over the same place of the planet at all times, it must be in a Geosynchronous orbit.
 Some data needed for the problems:
 The radius of Earth is \begin{align*}6.4 \times 10^6 \ m\end{align*}
 The mass of Earth is about \begin{align*}6.0 \times 10^{24} \ kg\end{align*}
 The mass of Sun is about \begin{align*}2.0 \times 10^{30} \ kg\end{align*}
 The EarthSun distance is about \begin{align*}1.5 \times 10^{11} \ m\end{align*}
 The EarthMoon distance is about \begin{align*}3.8 \times 10^8 \ m\end{align*}
Gravity Problem Set
 Which is greater – the gravitational force that the Earth exerts on the moon, or the force the moon exerts on the Earth? Why doesn’t the moon fall into the earth?
 Which is greater – the gravitational force that the Sun exerts on the moon, or the force the Earth exerts on the moon? Does the moon orbit the Earth or the Sun? Explain.
 Suppose you’re standing in an elevator on a bathroom scale. Draw a FBD for you and label the two forces acting on you. Describe how the scale reading compares to your weight when
 The elevator is at rest
 The elevator is moving up at a constant speed
 The elevator is accelerating upward
 The cable breaks and the elevator is in free fall
 Astronauts in orbit are not beyond the pull of earth’s gravity – in fact, gravity is what KEEPS them in orbit! So why are they “weightless”? Explain briefly.
 Use Newton’s Law of Universal Gravitation to explain why even though Jupiter has 300 times the mass of the earth, on the “surface” of Jupiter you’d weigh only 3 times what you weigh on earth. What other factor has to be considered here?
 Suppose we drilled a hole through the earth and dropped in a rock. How would the speed and acceleration of the rock vary as the rock falls into the earth? Would the rock come shooting out of the other side of the earth, or would it oscillate back and forth? Use a diagram in your explanation.
 There are basically three possible fates for a star that reaches the end of its life: white dwarf, neutron star, or black hole. What’s the main factor that determines which one a star will become? Which will our sun become? Explain what a black hole is. You may have to do a little research for this one.
 Do some research online to answer this question: there are three huge pieces of evidence for the Big Bang. Two cosmological phenomena called the cosmic microwave background radiation and the expansion of the universe. The third is called nucleosynthesis. Explain/define each, and briefly discuss how each supports the Big Bang theory. How do astronomers know the universe is expanding?
 Prove \begin{align*}g\end{align*} is approximately \begin{align*}10 \ m/s^2\end{align*}on Earth by following these steps:
 Calculate the force of gravity between a falling object (for example an apple) and that of Earth. Use the symbol \begin{align*}m_o\end{align*} to represent the mass of the falling object.
 Now divide that force by the object’s mass to find the acceleration \begin{align*}g\end{align*} of the object.
 Calculate the force of gravity between the Sun and the Earth. (sun mass \begin{align*}= 2.0 \times 10^{30} \ kg;\end{align*} average distance from sun to earth = 150 million km)
 Calculate the force of gravity between two human beings, assuming that each has a mass of 80 kg and that they are standing 1 m apart. Is this a large force?
 What is the force of gravity between an electron and a proton? Do you think this is what keeps the electron ‘orbiting’ the proton in the hydrogen atom?
 Calculate the gravitational force that your pencil or pen pulls on you. Use the center of your chest as the center of mass (and thus the mark for the distance measurement) and estimate all masses and distances.
 If there were no other forces present, what would your acceleration be towards your pencil? Is this a large or small acceleration?
 Why, in fact, doesn’t your pencil accelerate towards you?
 Mo and Jo have been traveling through the galaxy for eons when they arrive at the planet Remulak. Wanting to measure the gravitational field strength of the planet they drop Mo’s lava lamp from the top deck of their spacecraft, collecting the velocitytime data shown below.
velocity (m/s)  time (s) 

0  0 
3.4  1.0 
7.0  2.0 
9.8  3.0 
14.0  4.0 
17.1  5.0 

 Plot a velocitytime graph using the axes above. Put numbers, labels and units on your axes. Then draw a bestfit line (use a ruler) and use that line to find the gravitational field strength of Remulak. Explain below how you did that.

 Mo and Jo go exploring and drop a rock into a deep canyon – it hits the ground in 8.4 s. How deep is the canyon?
 If the rock has a mass of 25 g and makes a hole in the ground 1.3 cm deep, what force does the ground exert to bring it to a stop?
 Mo and Jo observe the shadows of their lava lamps at different positions on the planet and determine (a la Eratosthenes, the Greek astronomer, around 200 B.C.) that the radius of Remulak is 4500 km. Use that and your result for \begin{align*}g\end{align*} to find the mass of Remulak.
 A neutron star has a mass of about 1.4 times the mass of our sun and is about 20 km in diameter. It’s the remains of a star several times more massive than our sun that has blown off its outer layers in a supernova explosion. (see #1 for mass of sun)
 Calculate the density of a neutron star. (density = mass/volume); volume of a sphere is \begin{align*}4/3 \pi R^3\end{align*}.
 How much would you weigh on the surface of a neutron star? How many times greater is this than your weight on earth?
 Calculate \begin{align*}``g''\end{align*}, gravitational acceleration, on the surface of a neutron star.
 If you dropped your iPod from a height of 2.2 m on the surface of a neutron star, how long would it take to hit the surface? Compare this to the time it would fall on earth.
 At what height above the earth would a 400kg weather satellite have to orbit in order to experience a gravitational force half as strong as that on the surface of the earth?
 The Moon’s period around the earth is 27.3 days and the distance from the earth to the moon is \begin{align*}3.84 \times 10^8 \ m\end{align*}. Using this information, calculate the mass of the Earth. (note: this calculation assumes a circular orbit. Is your answer high or low? Why?).
 Calculate the mass of the Earth using only:
 Newton’s Universal Law of Gravity;
 the MoonEarth distance (\begin{align*}3.84 \times 10^8 \ m\end{align*}); and
 the fact that it takes the Moon 27.3 days to orbit the Earth.
Answers Problems:
 (c) \begin{align*}3.6 \times 10^{22} \ N\end{align*}
 \begin{align*}4.3 \times 10^{7} \ N\end{align*}
 About \begin{align*} 10^{47} N \end{align*}, No it is clearly not gravity.
(all answers are approximate for #13)

 \begin{align*}\sim 3.4 \ m/s^2\end{align*}
 \begin{align*}\sim 120 \ m\end{align*}
 \begin{align*}\sim 800 \ N\end{align*}
 \begin{align*}\sim 10^{23} \ kg\end{align*}

 \begin{align*}6.7 \times 10^{17} \ kg/m^3\end{align*}
 approx. \begin{align*}10^{14} \ N\end{align*}
 \begin{align*}1.9 \times 10^{12} \ m/s^2\end{align*}
 \begin{align*}1.5 \times 10^{6}s , 440,000\end{align*} times faster than on earth!
 \begin{align*}2.5 \times 10^6 \ m\end{align*} above earth’s surface
 Earth’s mass is about \begin{align*}6.0 \times 10^{24} \ N\end{align*}