I don't understand where g=6.667*10^-11N*m^2/kg^2 comes from.Can you help me understand how the book came up with that equation? 9.2 The Universal Gravitational Constant, GThe proportionality form of the universal law of gravitation can be expressed as anexact equation when the constant of proportionality G is introduced. G is calledthe universal gravitational constant. Then the equation isThe units of G make the force come out in newtons. The magnitude of G is thesame as the gravitational force between two 1-kilogram masses that are 1 meterapart: 0.0000000000667 newton. SoG = 6.67 X 10-¹¹N•m²/kg²This is an extremely small number.³ It shows that gravity is a very weak force com-pared with electrical forces. The large net gravitational force we feel as weight isbecause of the enormous number of atoms in planet Earth that are pulling on us.Interestingly, Newton could calculate the product of G and Earth's mass, butnot either one alone. G alone was first calculated by the English physicist HenryCavendish in 1798, more than 70 years after Newton's death.m1m2F = G-d²m1m₂d²Cavendish measured G by measuring the tiny force between lead masses with an ex-tremely sensitive torsion balance, as Professor Brage shows in one of the opening photosat the beginning of this chapter. A simpler method was later developed by Philipp vonJolly, who attached a spherical flask of mercury to one arm of a sensitive balance (Fig-ure 9.4). After the balance was put in equilibrium, a 6-ton lead sphere was rolled be-neath the mercury flask. The gravitational force between the two masses was measuredby the weight needed on the opposite end of the balance to restore equilibrium. All thequantities, m₁, m₂, F, and d, were known, from which the constant G was calculated:FG =6.67 × 10-¹1 N/kg/m² = 6.67 x 10-¹¹ Nm²/kg²-2yFIGURE 9.3As the rocket getthe gravitationalother decreases.Just as sheet rcian playing ma physics stucstand how caVIDEO:of MeasBetw

Gravitational force is attractive in nature. It is directly proportional to product of two masses…

I don't understand where g=6.667*10^-11N*m^2/kg^2 comes from. Can you help me understand how the book came up with that equatio

I don't understand where g=6.66710^-11Nm^2/kg^2 comes from. Can you help me understand how the book came up with that equatio

Physics for Scientists and Engineers: Foundations and Connections

1st Edition

ISBN:9781133939146

Author:Katz, Debora M.

Publisher:Katz, Debora M.

Chapter7: Gravity

Section: Chapter Questions

Problem 14PQ: Since 1995, hundreds of extrasolar planets have been discovered. There is the exciting possibility...

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Similar questions

The gravitational force exerted by the earth on a unit massat a distance r from the center of the planet is where M is the mass of the earth,R is its radius, and G isthe gravitational constant. Is F a continuous function of r ?
the standard acceleration (at sea level and 45 degree latitude) due to gravity is 9.80665 m/s^2. What isa at the force needed to hold a mass of 2 kg at rest in this gravitational field? How much mass can a force of 1N support?
In introductory physics laboratories, a typicalCavendish balance for measuring the gravitational constant G uses lead spheres of masses1.68 kg and 14.4 g whose centers are separatedby 2.7 cm.Calculate the gravitational force betweenthese spheres, treating each as a point masslocated at the center of the sphere. Thevalue of the universal gravitational constantis 6.67259 × 10−11 N · m2/kg2.Answer in units of N.
what is The gravitational constant? (I need Long answer plz)?
Here, (G = 6.67×10−11N m2/kg2) is the universal gravitational constant, (M) is the mass of the object,and (r) is its radius. For example, the mass of the Earth is (M = 6×1024kg) and the radius is (r = 6.4×106m). Thus, the surface gravity of Earth is:g=(6.67×10−11×6×1024(6.4×106)2)m/s2= 9.8 m/s2ObjectMassRadiusMercury3.3×1023kg2.4×106mVenus4.9×1024kg6.1×106mMars6.4×1023kg3.4×106mJupiter1.9×1027kg7.0×107mSaturn5.7×1026kg5.8×107mUranus8.7×1025kg2.5×107mNeptune1.0×1025kg2.5×107m ProcedureFor each of the planets listed above, compute the surface gravity in m/s21
Since 1995, hundreds of extrasolar planets have been discovered. There is the exciting possibility that there is life on one or more of these planets. To support life similar to that on the Earth, the planet must have liquid water. For an Earth-like planet orbiting a star like the Sun, this requirement means that the planet must be within a habitable zone of 0.9 AU to 1.4 AU from the star. The semimajor axis of an extrasolar planet is inferred from its period. What range in periods corresponds to the habitable zone for an Earth-like Planet orbiting a Sun-like star?
Compute directly the gravitational force on a unit mass at a point exterior to a homogeneous sphere of matter.
In introductory physics laboratories, a typical Cavendish balance for measuring the gravitational constant G uses lead spheres with masses of 1.90 kg and 19.0 g whose centers are separated by about 2.60 cm. Calculate the gravitational force between these spheres, treating each as a particle located at the center of the sphere.
In 2015, the U.S. national debt was about $18 trillion.(a) If payments were made at the rate of $1 000 per second,how many years would it take to pay off the debt, assumingthat no interest were charged? (b) A dollar bill is about 15.5cm long. If 18 trillion dollar bills were laid end to end aroundthe Earth’s equator, how many times would they encirclethe planet? Take the radius of the Earth at the equator to be6 378 km. (Note: Before doing any of these calculations, try toguess at the answers. You may be very surprised.)
An example of an inverse square field is the gravita- tional field F (mMG)r/|r|3 discussed in Example 4 in Section 13.1. Use part (a) to find thework done by the gravitational field when the earth moves from aphelion (at a maximum distance of 1.52x 102 km from the sun) to perihelion (at a minimum distance of 1.47x 102 (Use the values m= 5.97 x 1024 kg m=1.99x1030 kg and G=6.67 x 10-11 N-m2/kg2.
Compared to the strength of the Earth’s gravity at its surface where r = Re and Re is the radius of the Earth, how much weaker is gravity at a distance of 5Re? 15Re?
What is the gravitational force between two masses (m1 = 4 Kilograms and m2 -= 8 Kilograms) that are separated by a distance of 17 e meters? Express your answer as a number (rounded to one significant figure) and then the exponent of 10...for instance, express 3 x 10? as 3 and 2, express 5.3 x 10° as 5.3 and -3