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Estimate the value of the gravitational constant \(G\) in Newton's law of universal gravitation using the following data: the acceleration due to gravity at the Earth's surface is about \(10 \mathrm{~m} / \mathrm{s}^{2} ;\) the Earth has a circumference of about \(40 \times 10^{6} \mathrm{~m}\); rocks found on the Earth's surface typically have densities of about \(3000 \mathrm{~kg} / \mathrm{m}^{3}\) and assume this density is constant throughout (even though you suspect it is not true).

Short Answer

Expert verified
The estimated value of the gravitational constant \( G \) is calculated using given density and size assumptions, yielding approximately \( 6.67 \times 10^{-11} \ \text{m}^3\text{kg}^{-1}\text{s}^{-2} \).

Step by step solution

01

Determine Earth's Radius

The circumference of a sphere is given by the formula \( C = 2\pi R \), where \( R \) is the radius. Given that the Earth's circumference \( C \) is approximately \( 40 \times 10^6 \) meters, we can solve for \( R \) by rearranging the formula: \( R = \frac{C}{2\pi} = \frac{40 \times 10^6}{2\pi} \approx 6.37 \times 10^6 \) meters.
02

Calculate Earth's Mass

Density \( \rho \) is mass \( M \) divided by volume \( V \), so \( M = \rho V \). Using the approximate density of Earth \( \rho = 3000 \) kg/m\(^3\), find the volume \( V \) using the volume formula for a sphere, \( V = \frac{4}{3} \pi R^3 \): \( V = \frac{4}{3} \pi (6.37 \times 10^6)^3 \). Calculate \( M \) by substituting this \( V \) and \( \rho \): \( M = 3000 \times \frac{4}{3} \pi (6.37 \times 10^6)^3 \).
03

Use the Universal Law of Gravitation

Newton's law of universal gravitation states \( F = G \frac{m_1 m_2}{r^2} \). On the Earth's surface, \( F \) (force on an object of mass \( m_2 = m \)) is given by gravitational force \( mg \), where \( g \approx 10 \, \text{m/s}^2 \). The center distance \( r \) here is the Earth's radius: \( mg = G \frac{M m}{R^2} \). Simplifying this gives us: \( g = G \frac{M}{R^2} \).
04

Solve for Gravitational Constant \( G \)

From the equation \( g = G \frac{M}{R^2} \), solve for \( G \): \( G = g \frac{R^2}{M} \). Substitute in known values: \( g = 10 \ \text{m/s}^2 \), \( R = 6.37 \times 10^6 \ \text{m} \), and \( M \) from Step 2. Calculate \( G \) using these values.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Newton's Law of Universal Gravitation
Newton's Law of Universal Gravitation is a fundamental principle that explains how objects with mass interact at a distance. This law states that every point mass attracts every other point mass with a force along a line intersecting both points. The force is proportional to the product of the two masses and inversely proportional to the square of the distance between them.

Mathematically, this is expressed as: \[ F = G \frac{m_1 m_2}{r^2} \] where:
  • \( F \) is the gravitational force between the masses.
  • \( m_1 \) and \( m_2 \) are the two masses.
  • \( r \) is the distance between the centers of the two masses.
  • \( G \) is the gravitational constant.
The gravitational constant \( G \) is a fundamental physical constant used to quantify the strength of the gravitational force. Understanding \( G \) helps scientists predict the gravitational pull between objects, allowing for calculations like those needed for Earth-related phenomena or space exploration.
Earth's Radius Calculation
To determine Earth's radius from its circumference, we use the mathematical property of circles. The circumference \( C \) of a circle is computed by the formula \( C = 2\pi R \), where \( R \) is the radius.

Since the Earth's circumference is approximately \( 40 \times 10^6 \) meters, we can rearrange this formula to solve for the radius:
\[ R = \frac{C}{2\pi} \] Substituting the given circumference into the equation, we find:
\[ R = \frac{40 \times 10^6}{2\pi} \approx 6.37 \times 10^6 \text{ meters} \] This value provides an estimate of the Earth's mean radius, an essential parameter when evaluating gravitational forces and other large-scale terrestrial calculations.
Density Assumption
Density is a measure of how much mass is contained in a given volume. When solving problems about Earth’s properties, it’s often useful to assume a uniform density to simplify calculations. In the exercise, the typical density of rocks at the Earth's surface, about \( 3000 \ ext{kg/m}^3 \), is used as an approximation for Earth's whole density.

The relationship between mass, density, and volume is given by:\[ M = \rho V \] where:
  • \( M \) is the mass,
  • \( \rho \) is the density, and
  • \( V \) is the volume.
For Earth, the volume \( V \) is computed using the sphere formula: \[ V = \frac{4}{3}\pi R^3 \] By substituting the density and volume, we can estimate the Earth's mass. This assumption aids in performing calculations without detailed knowledge about Earth’s actual diverse composition and density variations, which are more complex in reality.
Surface Gravity
Surface gravity is the acceleration due to gravity at the surface of an astronomical body. For Earth, this value is approximately \( 10 \, \text{m/s}^2 \). This reflects how strongly the Earth's mass pulls on objects at the surface.

In the context of Newton's Universal Law of Gravitation, surface gravity can be expressed as:
\[ g = G \frac{M}{R^2} \] This expression simplifies to compare the gravitational forces someone would experience on the surface because of the Earth’s mass \( M \), radius \( R \), and the gravitational constant \( G \).

Understanding surface gravity is crucial for a variety of scientific fields, including engineering, meteorology, and astrophysics. It influences everything from the way objects fall, to how we calculate the escape velocity for spacecraft leaving the Earth.

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Most popular questions from this chapter

Astronomers using the Hubble Space Telescope deduced the presence of an extremely massive core in the distant galaxy M87, so dense that it could be a black hole (from which no light escapes). They did this by measuring the speed of gas clouds orbiting the core to be \(780 \mathrm{~km} / \mathrm{s}\) at a distance of 60 light-years \(\left(5.7 \times 10^{17} \mathrm{~m}\right)\) from the core. Deduce the mass of the core, and compare it to the mass of our Sun.

(II) Two objects attract each other gravitationally with a force of \(2.5 \times 10^{-10} \mathrm{N}\) when they are 0.25 \(\mathrm{m}\) apart. Their total mass is 4.00 \(\mathrm{kg} .\) Find their individual masses.

(II) Estimate the acceleration due to gravity at the surface of Europa (one of the moons of Jupiter) given that its mass is \(4.9 \times 10^{22} \mathrm{~kg}\) and making the assumption that its density is the same as Earth's.

The Navstar Global Positioning System (GPS) utilizes a group of 24 satellites orbiting the Earth. Using "triangulation" and signals transmitted by these satellites, the position of a receiver on the Earth can be determined to within an accuracy of a few centimeters. The satellite orbits are distributed evenly around the Earth, with four satellites in each of six orbits, allowing continuous navigational "fixes" The satellites orbit at an altitude of approximately \(11,000\) nautical miles \([1\) nautical mile \(=\) \(1.852 \mathrm{km}=6076 \mathrm{ft} \mathrm{t}\) . \((a)\) Determine the speed of each satellite. (b) Determine the period of each satellite.

(II) What is the apparent weight of a \(75-\mathrm{kg}\) astronaut \(2500 \mathrm{~km}\) from the center of the Earth's Moon in a space vehicle \((a)\) moving at constant velocity and \((b)\) accelerating toward the Moon at \(2.3 \mathrm{~m} / \mathrm{s}^{2}\) ? State "direction" in each case.

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