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Today’s APODAPOD  Start Reading NASA website (Oncourse)  2 nd Homework due TODAY  IN-CLASS QUIZ NEXT FRIDAY!! The Sun Today A100 Solar System

 Kepler's Laws were a revolution in regards to understanding planetary motion, but there was no explanation why they worked  The explanation was provided by Isaac Newton when formulated his laws of motion and gravity  Newton recognized that the same physical laws applied both on Earth and in space.  And don’t forget the calculus! Isaac Newton

Remember the Definitions Force: the push or pull on an object that affects its motion Weight: the force which pulls you toward the center of the Earth (or any other body) Inertia: the tendency of an object to keep moving at the same speed and in the same direction Mass: the amount of matter an object has

Newton's First Law  An object at rest will remain at rest, an object in uniform motion will stay in motion - UNLESS acted upon by an outside force Outside Force

Newton's Second Law  When a force acts on a body, the resulting acceleration is equal to the force divided by the object's mass  Acceleration – a change in speed or direction  Notice how this equation works:  The bigger the force, the larger the acceleration  The smaller the mass, the larger the acceleration or

Newton's Third Law  For every action, there is an equal and opposite reaction  Simply put, if body A exerts a force on body B, body B will react with a force that is equal in magnitude but opposite direction  This will be important in astronomy in terms of gravity  The Sun pulls on the Earth and the Earth pulls on the Sun

Newton and the Apple - Gravity Newton realized that there must be some force governing the motion of the planets around the Sun Amazingly, Newton was able to connect the motion of the planets to motions here on Earth through gravity Gravity is the attractive force two objects place upon one another

The Gravitational Force G is the gravitational constant G = 6.67 x 10 -11 N m 2 /kg 2 m 1 and m 2 are the masses of the two bodies in question r is the distance between the two bodies

Determining the Mass of the Sun  How do we determine the mass of the Sun?  Put the Sun on a scale and determine its weight???  Since gravity depends on the masses of both objects, we can look at how strongly the Sun attracts the Earth  The Sun’s gravitational attraction keeps the Earth going around the Sun, rather than the Earth going straight off into space  By looking at how fast the Earth orbits the Sun at its distance from the Sun, we can get the mass of the Sun

Measuring Mass with Newton’s Laws - Assumptions to Simplify the Calculation  Assume a small mass object orbits around a much more massive object  The Earth around the Sun  The Moon around the Earth  Charon around Pluto  Assume the orbit of the small mass is a circle

Measuring the Mass of the Sun  The Sun’s gravity is the force that acts on the Earth to keep it moving in a circle  M Sun is the mass of the Sun in kilograms  M Earth is the mass of the Earth in kilograms  r is the radius of the Earth’s orbit in kilometers  The acceleration of the Earth in orbit is given by: a = v 2 /r  where v is the Earth’s orbital speed

Measuring the Mass of the Sun  Set F = m Earth v 2 /r equal to F = GM Sun m Earth /r 2 and solve for M Sun M Sun = (v 2 r)/G  The Earth’s orbital speed (v) can be expressed as the circumference of the Earth’s orbit divided by its orbital period: v = 2  r/P

Measuring the Mass of the Sun  Combining these last two equations: M Sun = (4  2 r 3 )/(GP 2 )  The radius and period of the Earth’s orbit are both known, G and π are constants, so the Sun’s mass can be calculated  This last equation in known as Kepler’s modified third law and is often used to calculate the mass of a large celestial object from the orbital period and radius of a much smaller mass

So what is the Mass of the Sun? M Sun = (4  2 r 3 )/(GP 2 ) r Earth = 1.5 x 10 11 meters P Earth = 3.16 x 10 7 seconds G = 6.67 x 10 -11 m 3 kg -1 s -2 Plugging in the numbers gives the mass of the Sun M Sun = 2 x 10 30 kg

What about the Earth?  The same formula gives the mass of the Earth  Using the orbit of the Moon:  r Moon = 3.84 x 10 5 km  P Moon = 27.322 days = 2.36 x 10 6 seconds  Mass of Earth is 6 x 10 24 kg M Earth = (4  2 r 3 )/(GP 2 )

How about Pluto?  The same formula gives the mass of Pluto, too  Using the orbit of Pluto’s moon Charon:  r Charon = 1.96 x 10 4 km  P Charon = 6.38 days = 5.5 x 10 5 seconds  Mass of Pluto is 1.29 x 10 22 kilograms M Pluto = (4  2 r 3 )/(GP 2 )

Orbits tell us Mass  We can measure the mass of any body that has an object in orbit around it  Planets, stars, asteroids  We just need to know how fast and how far away something is that goes around that object  But we can’t determine the mass of the Moon by watching it go around the Earth  To determine the mass of the Moon, we need a satellite orbiting the Moon

Surface Gravity  Surface gravity:  Determines the weight of a mass at a celestial object’s surface  Influences the shape of celestial objects  Influences whether or not a celestial object has an atmosphere  Surface gravity is the acceleration a mass undergoes at the surface of a celestial object (e.g., an asteroid, planet, or star)

Surface Gravity Calculations  Surface gravity is determined from Newton’s Second Law and the Law of Gravity: ma = GMm/R 2  where M and R are the mass and radius of the celestial object, and m is the mass of the object whose acceleration a we wish to know  The surface gravity, denoted by g, is then: g = GM/R 2  Notice dependence of g on M and R, but not m  g Earth = 9.8 m/s 2  g Moon /g Earth = 0.18  g Jupiter /g Earth = 3

Escape Velocity To overcome a celestial object’s gravitational force and escape into space, a mass must obtain a critical speed called the escape velocity

Escape Velocity  Determines if a spacecraft can move from one planet to another  Influences whether or not a celestial object has an atmosphere  Relates to the nature of black holes

Escape Velocity Calculation  The escape velocity, V esc, is determined from Newton’s laws of motion and the Law of Gravity and is given by: V esc = (2GM/R) 1/2  where M and R are the mass and radius of the celestial object from which the mass wishes to escape  Notice dependence of V esc on M and R, but not m  V esc,Earth = 11 km/s, V esc,Moon = 2.4 km/s

Revisions to Kepler's 1st Law Newton's law of gravity required some slight modifications to Kepler's laws Instead of a planet rotating around the center of the Sun, it actually rotates around the center of mass of the two bodies Each body makes a small elliptical orbit, but the Sun's orbit is much much smaller than the Earth's because it is so much more massive

Revisions to Kepler's 3rd Law  Gravity also requires a slight modification to Kepler's 3rd Law  The sum of the masses of the two bodies is now included in the equation  For this equation to work, the masses must be in units of solar mass (usually written as M  )  Why did this equation work before?  Remember - for this equation to work:  P must be in years!  a must be in A.U.  M 1 and M 2 must be in solar masses

TO DO LIST:  Start Reading NASA website (Oncourse)  Hand in 2 nd Homework TODAY  IN-CLASS QUIZ NEXT FRIDAY!!

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