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Essential idea: The Newtonian idea of gravitational force acting between two spherical bodies and the laws of mechanics create a model that can be used to calculate the motion of planets.
Nature of science:
Laws: Newton’s law of gravitation and the laws of mechanics are the foundation for deterministic classical physics. These can be used to make predictions but do not explain why the observed phenomena exist. (2.4)
Understandings:
Newton’s law of gravitation
Gravitational field strength
Applications and skills:
Describing the relationship between gravitational force and centripetal force
Applying Newton’s law of gravitation to the motion of an object in circular orbit around a point mass
Solving problems involving gravitational force, gravitational field strength, orbital speed and orbital period
Determining the resultant gravitational field strength due to two bodies
Guidance:
Newton’s law of gravitation should be extended to spherical masses of uniform density by assuming that their mass is concentrated at their centre
Gravitational field strength at a point is the force per unit mass experienced by a small point mass at that point
Calculations of the resultant gravitational field strength due to two bodies will be restricted to points along the straight line joining the bodies
Data booklet reference:
Theory of knowledge:
The laws of mechanics along with the law of gravitation create the deterministic nature of classical physics. Are classical physics and modern physics compatible? Do other areas of knowledge also have a similar division between classical and modern in their historical development?
Utilization:
The law of gravitation is essential in describing the motion of satellites, planets, moons and entire galaxies
Comparison to Coulomb’s law (see Physics sub-topic 5.1)
Aims:
Aim 4: the theory of gravitation when combined and synthesized with the rest of the laws of mechanics allows detailed predictions about the future position and motion of planets
Newton's Law of Universal Gravitation - Two Bodies Interacting
Newton's Law of Universal Gravitation - Two Bodies Interacting
Sample Problems - Force Gravity
Sample Problems - Force Gravity
D = 1.33, (A,B,F) = 1, E = 0.75, C = 0.5
3. How far above the surface of the earth must you be for your weight to be only 1/4 of what it is when you are on the earth's surface? (RE = radius of the earth)
RE ✔
2 RE
3 RE
7 RE
15 RE
5. Calculate the gravitational force of attraction between two metal spheres each of mass 90 kg, if the distance between their centres is 40 cm. Given G = 6.67 x 10-11 N m2/kg2. Will the force of attraction be different if the same bodies are taken on the moon, their separation remaining the same?
6.
Q2: Satellites #1 and #2 have the same mass and both orbit earth in concentric circles. Satellite #2 is twice as far from earth's center as satellite #1.What is the ratio of the gravitational force acting on #2 to that acting on #1?
1/8
1/4
1/2 ✔
1/√ 2
1 ✔
4. An astronaut has a mass of 72 kg on Earth. He lands on the moon where the gravitational field strength is 1.6 N kg-1.
State the mass of the astronaut on the moon. 72 kg
Calculate the weight of the astronaut
on the Earth, 706.3 N
on the moon. 115.2 N
Determine the Gravitational Forces between the following objects.
Gravitational Field Strength - Property of Single Body
Gravitational Field Strength - Property of Single Body
Show that the following equations are equivalent:
The gravitational field strength at a point is the force per unit mass experienced by a test mass at that point.
g = gravitational field strength (Effect)
F = Force Gravity (Cause)
m = mass of test object (not the central body) (Limiter)
Gravitational field strength at the surface of a planet
The gravitational field strength at the surface of a planet can be calculated by using the equation for gravitational field strength and substituting M and r by the mass and the radius of the planet respectively.
If we calculate the gravitational field strength at the surface of the Each using the mass and the radius of the Earth, we would obtain the value 9.81m/s^2, which is equal to the acceleration due to gravity on the surface of the Earth.
Different planets have different radii and masses. Consequently, different planets have different gravitational field strengths.
Sample Problems - Gravitational Field Strength
Sample Problems - Gravitational Field Strength
2. Gravitational field strength at a point may be defined as
A. the force on a small mass placed at the point.
B. the force per unit mass on a small mass placed at the point. ✔
C. the work done to move unit mass from infinity to the point.
D. the work done per unit mass to move a small mass from infinity to the point.
3. A space probe of mass 5.60 x 10^3 kg is in orbit at a height of 3.70 x 10^6 m above the surface of Mars.
The mass of Mars is 6.42 x 10^23 kg. The radius of Mars is 3.39 x 10^6 m.
(a) Calculate the gravitational force between the probe and Mars. 4770 N
(b) Determine the velocity of the space probe at this elevation. 2457 m/s
(c) Calculate the gravitational field strength of Mars at this height. 0.85 N/kg
4. Planet X has radius R and mass M. Planet Y has radius 2R and mass 8M.
Which one of the following is the correct value of the ratio
A. 4 B. 2
C. 1/2 ✔ D. 1/4
5. An object of mass m released from rest near the surface of a planet has an initial acceleration z. What is the gravitational field strength near the surface of the planet?
A. z ✓ B. z/m C. mz D. m/z
Circular Motion Problem:
Circular Motion Problem:
6. DAWN Space Probe: The NASA space probe Dawn has travelled to and orbited large asteroids in the solar system. Dawn has a mass of 1240 kg.
One particular asteroid that Dawn orbited was Vesta. It can be considered to be spherical and remote from other large objects.
The mass of Vesta is 2.59 x 10^20 kg. The radius of Vesta is 263 km.
Dawn began orbiting Vesta, in a circular orbit, at a height of 680 km above the surface of the asteroid.
Determine the gravitational field strength of the asteroid Vesta at the orbital height of DAWN. g = 0.0194 N/kg
Compare the field strength of Vesta to that of the surface of the Earth. 0.00198xEarth's
Determine the force acting between the asteroid and space probe at an orbital distance of 680 km. F = 24 N
Show that the tangential velocity of Dawn in this orbit is 135 m s-1.
Calculate the orbital period of Dawn. T = 43889s ~ 12h12min
Gravitational Potential Energy - Ug
Gravitational Potential Energy - Ug
Gravitational P.E. is negative at the surface of Earth, because work is completed by the field in bringing a mass from infinity i.e., work needs to be done on a body, if it's moved out from the field of the earth. Thus, P.E. is negative.
As expected, all values of the potential energy are negative, approaching the value of zero as R approaches infinity. Because of this we say that the mass is trapped in an "energy well" - that is, it will have to be given additional energy from an external source if it is to escape gravity's attraction.
Q1: A 10-kg satellite is orbiting 36 km above the surface of the earth in circular orbit. ME = 5.98 x 1024 kg and rE = 6.37 x 106 m
What is the satelite's gravitational potential energy? Ans: Ug = -6.22 x 108 J
What is the satellite's kinetic energy? Ans: v = 7890 m/s K = 3.11 x 108 J
What is the satellite's total energy? Ans: ET = -3.11 x 108 J
What is the significance of the negative sign on your previous question?
Q2: NASA, the ESA and the UAE Space Agency need to move the International Space Station (ISS) to a higher orbit to UAE's astronaut Sultan AlNeyadi perform vital research in a micro-gravity environment. The ISS has a mass of 420 000 kg. The Space agencies would like to move the ISS from an orbit of 420 km to an orbit of 550 km. ME = 5.98 x 1024 kg and rE = 6.37 x 106 m
Determine the force of gravity acting on the ISS at an orbital height of 420 km. Ans: ME = 3.63 x 106 N
Determine the tangential velocity of the ISS at an orbit height of 420 km. Ans: 7664 m/s
Calculate the orbital period of the ISS. Ans: 5566 s ≅ 90 min
Determine the energy necessary to move the ISS to its new orbital height. (ΔUg ) Ans: Ug-420km = -2.46 x 1013 J , Ug-550km = -2.42 x 1013 J ∆Ug = 4.36 x 1011 J
Outline if the velocity will be faster or slower at the new height of 550 km. Ans: Slower
Rocket fuel contains 43 MJ/kg of energy. How many kilograms of fuel must be burned to move the ISS to this new orbital height. Ans: 10 800 kg
Where does the r go?
Where does the r go?
The force gravity equation is based upon the inverse square of the radius, where as the potential energy equation is on dependent upon the inverse of the radius.
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