Newton's Law of Universal Gravitation

2 G=6.67x10 distance

Gravitational Field Strength


Gravitational Field of a Hollow Shell

  • Inside a hollow sphere, the gravitational field is 0. Outside a hollow sphere, you can treat the sphere as if it's entire mass was concentrated at the center, and then calculate the gravitational field


Gravitational Field Inside a Solid Shell

  • Outside a solid sphere, treat the sphere as if all the mass is at the center of the sphere. Inside the sphere, treat the sphere as if the mass inside the radius is all at the center. Only the mass inside the "radius of interest" counts


Velocity in Circular Orbit


Period and Frequency for Circular Orbits


Mechanical Energy for Circular Orbits


Escape Velocity


Kepler's Frist Law of Planetary Motion

  • The orbits of planetary bodies are ellipses with the sun at one of two foci of the ellipse

    01 íd un” saaouj łauełd Z eauv = L eauv eaJV d ld

Kepler's Second Law of Planetary Motion

  • If you were to draw a line from the sun to the orbiting body, the body would sweep out equal areas along the ellipse in equal amounts of time.

    Pd 01 Ed laueld Z eaJV = I eaav d (uns)

Kepler's Third Law of Planetary Motion

  • The ratio of the squares of the periods of two planets is equal to the ratio of the cubes of their semi-major axes.

  • The ratio of the squares of the periods to the cubes of their semi-major axes is referred to as Kepler's Constant

    Perigee Point B semi-minor axis (b) semi-ma or axis (a) Apogee Point A

Total Mechanical Energy for an Elliptical Orbit


Velocity and Radius for an Elliptical Orbit

perigee Point B p X ve be semi-minor axis (b) semi-ma•or axis (a) Point A Apogee

Example 1: Rocket Launched Vertically

  • A rocket is launched vertically form 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 (m1) is 6*1024</sup kg and the radius of the Earth (RE) is 6.37*106</sup m. You may not assume that the acceleration due to gravity is constant.

    K; لا : اوكا أس - سيس؟ ى ح مدي -ل يا ادت ا .Pـ r 10• مه (ى)(6"c)2 10•6.37 -٧ -٣ ٣٤3١2 ت٥/: - ٠.6.37٥2

2007 Free Response Question 2

Mech. 2. In March 1999 the Mars Global Surveyor (GS) entered its final orbit about Mars, sending data back to Earth. Assume a circular orbit with a period of 1.18 X 102 minutes = 7.08 X 103 s and orbital speed of 3.40 X 103 m/s . The mass of the GS is 930 kg and the radius of Mars is 3.43 X 106 m . (a) Calculate the radius of the GS orbit. (b) Calculate the mass of Mars. (c) Calculate the total mechanical energy of the GS in this orbit. (d) If the GS was to be placed in a lower circular orbit (closer to the surface of Mars), would the new orbital period of the GS be greater than or less than the given period? Greater than Justify your answer. Less than (e) In fact, the orbit the GS entered was slightly elliptical with its closest approach to Mars at 3.71 X 105 m above the surface and its furthest distance at 4.36 X 105 m above the surface. If the speed of the GS at closest approach is 3.40 X 103 m/s , calculate the speed at the furthest point of the orbit.

zry z

、 0 4 。 ハ い 。 「 、 ← 0 」 ト 次 3 ( 「 , ト ー も 。 、 ミ ぶ 0

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