What affects the gravitational attraction between two objects?
Physics 1 Review Stuff
▪ Weight – Force exerted on an object by the earth or another body.
▪ Newton’s Law of Gravitation – Every object in the universe attracts every other particle:
*The direction of the gravity force is along a straight line connecting the center of mass of the objects.
▪ Weight on earth is mg, but it is also the force of gravity between an object and the planet its own:
▪
Using the two formulas for weight (mg and the force of gravity):
Acceleration due to gravity on a given: 
▪ M – mass of planet r – radius of planet g – acceleration due to gravity on planet
Example A: Planet X has half the radius and half the density of Earth. What is the acceleration due to gravity on planet X?
Answer/Explanation
Ans: When given density and radius, a new formula needs to be derived to relate acceleration due to gravity to density. Using the formula for acceleration due to gravity, mass can be replaced by using the fact that density (𝜌) is mass over volume. A planet is roughly spherical, so the volume for a sphere is used:
\(g=\frac{GM}{r^{2}}=\frac{G\rho V}{r^{2}}=\frac{G\rho \frac{4}{3}\pi r^{3}}{r^{2}}\Rightarrow g=G\rho \frac{4}{3}\pi r\)
Now, ratios can be used again to find the acceleration on the unknown planet.
\(\frac{gx}{gE}=\frac{G\rho x\frac{4}{3}\pi rx}{G\rho E\frac{4}{3}\pi rE}=\frac{\frac{1}{2}\rho E\frac{1}{2}rE}{\rho ErE}=\frac{\frac{1}{4}}{1}\)
\(\Rightarrow gx=\frac{1}{4}gE\Rightarrow gx = 2.5 m/s^{2}\)
Kepler’s Laws
▪ Kepler’s First Law
Concept: Planets have elliptical orbits with its star as a focus.
Math: Average orbital speed can be calculating using circular motion.
▪ If a satellite has a circular orbit with constant velocity, then the satellite experiences an acceleration \(\frac{v^{2}}{r}\) directed toward the center of the orbit. The force providing the force to keep it in its orbit is gravity:
▪ Kepler’s Second Law:
Concept: Planets move faster closer to their star.
Math: Angular momentum of a planet about its star is conserved.
“Moment of inertia” of planet about star: 𝑰 = 𝒎𝒓
Angular momentum of planet about star:𝑳 = 𝒎𝒗𝒓
*Angular momentum of planet about its star is always conserved during a planetary orbit since no outside forces act on the planet/star system.
𝒎𝒗𝟏𝒓𝟏 = 𝒎𝒗𝟐𝒓𝟐
▪ Kepler’s Third Law:
The square of orbital period is proportional to the cube of average orbital radius.
▪ Gravitational Potential Energy (with respect to the universe) – The potential energy between object on a planet and the planet is equal in magnitude to the work done to bring it equal from its current location to a point infinitely far away in space where the potential energy due to the planet is zero:
Potential energy is equal in magnitude to the work required to give something potential energy:
This energy is negative. This is because it would take positive work just to reach a point in space with no energy. Potential energy is the ability to do work. Since it takes work to have no energy, there is a large amount of negative energy.
* This is technically true for any two objects with mass, but is only significant for celestial bodies. It takes energy to separate two bodies that are bound together with gravity.
Mechanical Energy for a Circular Orbit
Total Mechanical Energy is the sum of kinetic and potential energies: A satellite in orbit has a large kinetic energy, but it’s overall energy is still negative since it would take work to escape planet’s gravity –
Example B: Matt Damon pilots a 6000 kg spaceship around Mars. Mars has a mass of 6.4×1023 kg and a radius of 3.4×106 m.
a) If Matt Damon orbits Mars in a circular orbit with at a distance of 6000 km above the surface of Mars, how long would it take him to complete one orbit of Mars? Answer in hours.
Answer/Explanation
Ans: 
\(\frac{mv^{2}}{r}=\frac{GMm}{r^{2}}\rightarrow v=\sqrt{\frac{GM}{r}}\)
\(T=\frac{2\pi r}{v}=\frac{2\pi r}{\sqrt{\frac{GM}{r}}}\)
\(\Rightarrow T=\frac{2\pi(3.4\times 10^{6}+6000\times 10^{3})}{\sqrt{\frac{(6.67\times 10^{-11})(6.4\times 10^{25}kg)}{(3.4\times 10^{6}+6000\times 10^{3})}}}=27716s\)
= 7.7 hours
b) Now, Matt Damon orbits Mars in an elliptical orbit, with Mars acting as one of the foci. The farthest point from Mar in the orbit is A and the closest is B. A is a distance of 30×106 m from the center of Mars and point B is 12×106 from the center of Mars.
i. Matt Damon’s rocket has a speed of 900 m/s at point A. Calculate its total mechanical energy at this point.
Answer/Explanation
Ans: \(E=\frac{-GMm}{r}+\frac{1}{2}mv^{2}=\frac{-GMm}{r}+\frac{1}{2}m\frac{GM}{r}=\frac{-GMm}{2r}\)
\(E=\frac{-(6.67\times 10^{-11})(6.4\times 10^{23}kg)(6000 kg)}{2(30\times 10^{6}m)}=-4.3 \times 10^{9}J\)
Even though Matt Damon has positive kinetic energy, his overall energy is still very negative as he is still within Mar’s gravity well.
ii. Describe how the speed of Matt Damon’s rocket will change as it moves towards point B.
Justify your answer.
Answer/Explanation
Ans: As Matt Damon flies closer to Mars, his speed will increase in order to conserve angular momentum (gravity is also stronger closer to the planet).
iii. Calculate the speed of the rocket at point B.
Answer/Explanation
Ans: \(MV_{A}r_{A}=MV_{B}r_{B}\Rightarrow V_{B}=\frac{V_{A}r_{A}}{r_{B}}=\frac{(900 m/s)(30\times 10^{6}m)}{12\times 10^{6}m}\)
⇒ VB = 2250 m/s
Example C: A rocket is launched vertically from the surface of the Earth with an initial velocity of 10 km/s.
What maximum height does it reach, neglecting air resistance? Note that the mass of the Earth is 6×1024 kg and the radius of the Earth is 6.37×106 m. You may not assume that the acceleration due to gravity is a constant
Answer/Explanation
Ans: This is solve using conservation of energy since kinematics cannot be used. h can be found with a solver.
\(U_{i}+k_{i}=U_{f}+k_{f}\Rightarrow \frac{-GMm}{r}+\frac{1}{2}mv^{2}=\frac{-GMm}{r+h}\)
\(\frac{6.67\times 10^{-11}(5.97\times 10^{24}kg)}{7.4\times 10^{6}m}+\frac{1}{2}(1000 m/s)^{2}=\frac{6.67\times 10^{-11}(5.97\times 10^{24}kg)}{6.4\times 10^{6}m+h}\)
Escape Velocity
To derive the escape velocity formula, conservation of energy will be used. For minimum escape velocity, the final energy will be zero as the object will have escape all negative potential energy gravity wells. When the object is launched, it has negative potential energy and positive kinetic energy.
Example D: Calculate the escape velocity for a 6000 kg spacecraft that wants to escape Earth’s gravity well.
Answer/Explanation
Ans: \(v = \sqrt{\frac{2(6.67\times 10^{-11}\frac{Nm^{2}}{kg})(5.97\times 10^{24}kg)}{6.4\times 10^{6}}m}=11155 m/s \approx 11 km/s\)
This means an object would have to be going 11 km/s at launch to escape Earth. This is if there is no force propelling the object while in the air.
Calculate gravitational fields inside and outside a solid shell of mass.
▪ The total gravitational field at a point is the vector sum of all gravitational fields of the individual masses that produce the field. If you know the field at a point, you can calculate the gravitational force at that point.
Example E: An object of mass m is in space at point P at a distance of D away from the center of a hollow ring of mass M. Find the force of gravity between the 2 masses.
Answer/Explanation
Ans: 
The y–components of gravity cancel out so only the x–component needs to be solve for as follows:
\(dF_{x}=\frac{GmdM}{r^{2}}cos\theta ,\)
▪ Gravitational field due to a hollow sphere.
Again, only the x–component is calculated here since y cancels out. The sphere is separated into a bunch of time rings of mass dM.
\(dF_{x}=\frac{GmdM}{s^{2}}cos\phi \)
where 𝑑𝑀 = 𝑑𝑒𝑛𝑠𝑖𝑡𝑦 𝑜𝑓 𝑟𝑖𝑛𝑔 𝑥 𝑎𝑟𝑒𝑎 𝑜𝑓 𝑟𝑖𝑛𝑔 → 𝑑𝑒𝑛𝑠𝑖𝑡𝑦 𝑥 𝑐𝑖𝑟𝑐𝑢𝑚𝑓𝑒𝑟𝑒𝑐𝑒 𝑥 𝑡ℎ𝑖𝑐𝑘𝑛𝑒𝑠𝑠
𝑑𝑀 = 𝜎(2𝜋𝑅 sin 𝜃)(𝑅𝑑𝜃)
\(\rightarrow dF_{x}=\frac{GmMsin\theta d\theta }{2s^{2}}cos\phi \)
This is bombad and very scary since 𝜃, 𝜙, and s are all variables, so some math is required.
Law of cosines for 𝜙: 𝑅2 = 𝑠2 + 𝑟2 − 2𝑠𝑟 cos 𝜃
Law of cosines for 𝜃: 𝑠2 = 𝑟2 + 𝑅2 − 2𝑟𝑅 cos 𝜃
Differentiating: 2𝑠𝑑𝑠 = 2𝑟𝑅 sin 𝜃 𝑑𝜃
\(\rightarrow sin\theta d\theta =\frac{sds}{Rr}\)
This can be plugged into the original equation to get everything in terms of the variable s:
▪ Gravitational field inside a hollow shell.
Anywhere inside a hollow shell will have gravity cancel out. Any mass attracting the object “above” it will cancel out with a piece of mass somewhere else inside the shell. Only mass “below” an object will attract a mass to a large body.
▪ Graph of Gravitational Field Starting at Earth’s Center
Since gravity field strength is only due to mass “under” an object, it will increase higher up in Earth’s crust. The maximum value will be at the surface, after which point, g will decrease asymptotically according to \(g=\frac{GM}{r^{2}}.\)
