Lecture 1 Overview of Physical Geodesy PDF

Summary

This lecture introduces fundamental concepts in physical geodesy, focusing on gravity, gravitation, and Newton's laws. It covers topics such as the definition of physical geodesy, gravitation (force of attraction between particles), and Newton's laws of motion. The lecture is aimed at undergraduate students.

Full Transcript

Overview of Physical Geodesy GSS611 Definition of Physical Geodesy Study of the physical properties of the gravity field of the earth, the geopotential , with a v...

Overview of Physical Geodesy GSS611 Definition of Physical Geodesy Study of the physical properties of the gravity field of the earth, the geopotential , with a view to their application on geodesy i. Geoid undulation ii. Gravimetric deflection iii. Earth’s shape and size (semi major, semi minor and flattening) What is gravity (gravity force)? Gravity is the force that pull you down or holds you direction and magnitude of the in place centrifugal effect Physics- Resultant force experienced on the net result of gravity plus centrifugal effect Earth’s surface due to the attraction by the earth’s gravitational field strength masses (gravitation), and the centrifugal force caused by the earth rotation Gravitation (Gravitational Force) The vector force of attraction that exist between all particles with mass in the universe (Newton’s 3rd Law) Responsible for holding objects onto the surface of planets and with Newton’s law inertia (Newton 1st Law) is responsible for keeping objects in orbit around another Describe by Newton’s Law of universal gravitation Newton’s Three Laws Newton’s 1st Law Also called the Law of Inertia or Galileo Principle Every object persists in its state of rest, or uniform motion (in a straight line); unless it is compelled to change that state, by force impressed on it A body remains at rest, or moves in a straight line (at a constant velocity), unless acted upon by a net outside force Newton’s Three Laws Newton’s 2nd Law The time rate of change in momentum is proportional to the applied force and takes place in the direction of the force The acceleration of an object is proportional to the force acting upon it This is expression by the equation : F=ma where F= force , m= mass and a=acceleration Newton’s Three Laws Newton’s 3rd Law Whenever one body exerts force upon a second body, the second body exerts an equal and opposite force upon the first body For every action, there is an equal and opposite reaction Newton’s Law of Universal Gravitation Based on Newton’s 3rd law “ Every object in the Universe attract every other object with a forced directed along the line of centers for the two objects that is proportional to the product of their masses and inversely proportional to the square of the separation between the two objects” F= Gravitational force between two objects m1,m2 = Mass of 1st and 2nd r = distance between the objects G= 6.674215 x 10-11 Nm2kg-2 Newton’s Law of Universal Gravitation Explain many natural phenomena: Plenatary motion Free fall Tides Equilibrium shape of the earth Fundamental observation on gravitation: The force between two attracting bodies is proportional to individual masses The force is inversely proportional to square of the distance The force is directed along the line connecting the two bodies Gravity field strength Based on the Newton’s 2nd and 3rd Law F=ma or F = mg F= Thus, if we simplify the formula m1g = g =GM / r2 Gravity field strength Gravitational field strength generated by the earth at the surface ? mass of earth, M = 5.97 X 1024 radius of the earth , r = 6380000m gravitational constant ,G = 6.67 x 10-11 Nm2 kg-2 g =GM / r2 = (6.67 x 10-11 x 5.97 X 1024 ) / 6380000 = 9.78 x N/kg = 9.81 ms-2 Gravity field strength Gravitational field strength generated by the earth at the 5000km above the surface ? mass of earth, M = 5.97 X 1024 radius of the earth , r = 6380000m gravitational constant ,G = 6.67 x 10-11 Nm2 kg-2 g =GM / r2 = (6.67 x 10-11 x 5.97 X 1024 ) / (6380000+5000000) = 3.07 ms-2 Gravity field strength Gravitational field strength generated by the earth at the 10000km above the surface ? g at 5000km = 3.07 m/s-2 g at 10000km = 3.07ms-2 / 4 Exercise Calculate the mass of the Earth if gravitational force, G is 6.67 x 10-11 m3kg-1s-2, the radius of the Earth, r is 6.37 x 106 m and gravity acceleration, g is 9.8 ms-2. m Mass of the Earth F = GmM/r2 = ma, where F is the gravitational force, G is the gravitational constant, M is the mass of the Earth, r is the radius of F=ma the Earth, and m is the mass of another object (near the surface of 𝐺𝑚𝑀 r the Earth). 𝐹= 𝑟2 GM/r2= a (The m is canceled out.) Now solve for M, the mass of the Earth. M = ar2/G M M = 9.8 x (6.37 x 106)2/(6.67 x 10-11) = 5.96182 x 1024 kg Centrifugal and Centripetal Force The gravitational force is balance by a reaction force known as centrifugal force It is actually not a force but the experience of an inertial force experienced in rotating reference frame acting away from the center of the rotation Equal in magnitude but opposite to the centripetal force required to constrain to body to move in circular motion Directed outward and is perpendicular to the rotation axis Centrifugal Force For a perfectly spherical shape, the force defined as 𝐹𝑐 = 𝑚1 𝜔2 𝑙 = 𝑚1 𝜔2 𝑅 𝑐𝑜𝑠 𝜙 𝐹𝑜𝑟 1 𝑢𝑛𝑖𝑡 𝑚𝑎𝑠𝑠, 𝑭𝒄 = 𝝎𝟐 𝑹 𝒄𝒐𝒔 𝝓 𝑊ℎ𝑒𝑟𝑒: 𝑚1 = 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑡ℎ𝑒 𝑝𝑜𝑖𝑛𝑡 𝑎𝑐𝑡𝑒𝑑 𝑢𝑝𝑜𝑛 𝜔 = 𝑎𝑛𝑔𝑢𝑙𝑎𝑟 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑜𝑓 𝑡ℎ𝑒 𝑟𝑜𝑡𝑎𝑡𝑖𝑛𝑔 𝑒𝑎𝑟𝑡ℎ = 7292115𝑥10−11 𝑟𝑎𝑑 𝑠 −1 𝑙 = 𝑟𝑜𝑡𝑎𝑡𝑖𝑜𝑛 𝑎𝑟𝑚 = 𝑅𝑐𝑜𝑠𝜙 (depend on latitude) At the pole and equator, the centrifugal force are; 𝑷𝒐𝒍𝒆 𝝓 = 𝟗𝟎𝒐 ∶ 𝐹 = 𝜔2 𝑅 𝑐𝑜𝑠 𝜙 = 𝜔2 𝑅 𝑐𝑜𝑠 90 = 0 𝑐 𝑬𝒒𝒖𝒂𝒕𝒐𝒓 𝝓 = 𝟎𝒐 ∶ 𝐹𝑐 = 𝜔2 𝑅 𝑐𝑜𝑠 𝜙= 𝜔2 𝑅 𝑐𝑜𝑠 0 = 𝜔2 𝑅 1 𝑇ℎ𝑖𝑠 𝑖𝑠 𝑜𝑛𝑙𝑦 𝑜𝑓 𝑔𝑟𝑎𝑣𝑖𝑡𝑦 300 Hence, maximum at equator and minimum at pole. Centripetal and Gravitational force Centripetal is a force that makes a body follow a curved path. Its direction is always orthogonal to the motion of the body and towards the fixed point of the instantaneous center of curvature of the path To make a body follow a curved path, the centripetal and gravitational should be same and balance with the speed of the body Centripetal and Gravitational force Fc = Fg v 𝑚𝑣 2 𝐺𝑀𝑚 = 𝑅 𝑅2 F Centripetal and Gravitational force Example If the mass of the earth is 5.97x 1024 kg and radius of the earth is 6.38 x106m, find the speed of a satellite moving in circular orbit at the height of 3800km above the surface of the earth and the period of the satellite in hours Satellite Earth Centripetal and Gravitational force The speed of a satellite moving in circular orbit Satellite Earth 𝑚𝑣 2 𝐺𝑀𝑚 = 𝑅 𝑅2 Simplify 𝐺𝑀 (6.67𝑥10−11 )(5.97x 1024) 𝑣 = 6254.3 𝑚/𝑠 𝑣= 𝑣= 𝑅 3.8 𝑥106 + 6.38 x106 Centripetal and Gravitational force The period of the satellite in hours Satellite Earth 𝑑 v= 𝑡 2𝜋𝑅 v= 𝑇 2𝜋𝑅 2𝜋3.8 𝑥106 + 6.38 x106 T= T= 𝑣 6254.3 T = 10227𝑠 ~ 2.84ℎ𝑜𝑢𝑟𝑠 Gravity The force of gravity at any point on the earth is the vectorial resultant of the gravitational force,𝐹 and centrifugal force,𝐹𝑐 In simplified form: 𝐺𝑀𝑚 𝑔 = 𝐹 + 𝐹𝑐 = 2 + 𝜔2 𝑅𝑐𝑜𝑠𝜙 𝑅 In vectorial resulatant (considering 3 sides of g,F, Fc) with a unit mass: 𝑔2 = 𝐹 2 + 𝐹𝑐2 − 2𝐹𝐹𝑐 cos 𝜙 (Cosine rule) 2 𝐺𝑀 2𝐺𝑀 2 𝑔2 = + (𝜔2 𝑅𝑐𝑜𝑠𝜙)2 − 𝜔 𝑅 cos 2 𝜙 𝑅2 𝑅 2 Hence: 𝑮𝑴 𝝎𝟐 𝑹𝟑 𝒈 ≈ 𝟐 𝟏− 𝒄𝒐𝒔𝟐 𝝓 𝑹 𝑮𝑴 Gravity The earth’s gravitational vector field can be represented as the gradient of a scalar function called gravitational potential (each field contain constant value of gravitational potential) Why gravity at pole is stronger than at equator?? At equator: Gravitational force and centrifugal force are along the same line but opposite in direction Simply added, gE ≈ 9.78ms-2 At pole: Gravitational force and centrifugal force are perpendicular to each other direction and magnitude of the Simply , gP ≈ 9.83ms-2 centrifugal effect net result of gravity plus centrifugal effect Gravitational Potential Energy (GPE) What is GPE? Energy stored in a mass due to its position in a gravitational field GPE The potential energy on the Earth’s surface at ground level is taken to be equal to 0 Thus, for energy changes in a uniform gravitational field (such as near the Earth’s surface), the GPE: Gravitational Potential Energy (GPE) ∞ GPE =max Gravity strength =0 Surface GPE=0 Gravitational Potential (V) What is GP? the work (energy transferred) per unit mass that would be needed to move an object to that location from a fixed reference location 𝑊 (Unit : Jkg-1) 𝑉=− 100kg 𝑚 𝐺𝑀 𝑉=− Earth surface 𝑟 Gravitational Potential (V) 100kg Earth surface 𝑤 = 𝑚∆𝑉 = 100 x (1.49 x 107 Jkg-1) = 1.49 x 109J Gravitational Potential (V) Earth surface R r ∆𝑉 −𝐺𝑀 =− / r ∆𝑟 𝑟 −𝐺𝑀 =− 2 𝑟 -V Gravitational Potential, V For various point masses m1, m2, m3,…,mn 𝑛 𝑚𝑖 𝑉= 𝐺 σ𝑖=1 𝑟 𝑖 For solid body such as earth Potential is a volume integration, ~ Triple integrals = volume Where: ~ dm : Element of mass outside or inside the sphere ~ dv : Element of volume 𝑑𝑚 𝜌𝑑𝑣 𝑉 = 𝐺ම = 𝐺ම 𝑟 𝑟 Equipotential Surface The imaginary surface on which at every point , the potential is same. Equipotential Surface Often called level surface and the perpendicular is known as the vertical or plumbline. Direction of gravity must be everywhere perpendicular to this surface as defined by plumbline. A carefully leveled theodolite defines horizontal and vertical planes. Covers the earth like layers of an onion (do riot cross and are parallels to each others). Continuous, having no sharp edges and are convex everywhere with smoothly varying radii of curvature. Sectional view: Shows oblate curves spaced closer together at poles than at equator with vertical as a curved lines intersecting each surface at right angles. Relationship between gravity and equipotential surface Direction of gravity and the equipotential surface are mutually perpendicular Spacing of the surface is directly related to the magnitude of gravity. Important of Equipotentional Surface Elevation Measured with respect to sea level, or orthometric height using levelling. Levelling Start at sea level and call this zero elevation. (If there were no winds, currents and tides then the ocean surface would be an equipotential surface and all shorelines would be at exactly the same potential.) Sight a line inland perpendicular to a plumb line (line will be perpendicular to the equipotential surface and thus is not pointed toward the geocenter). Measure the height difference and then move the setup inland and to repeat the measurements until reaching the next shoreline (zero elevation is obtained assuming the ocean surface is an equipotential surface and no errors in observation).

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