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Chapter 6: Problem 6

(II) Calculate the effective value of \(g\), the acceleration of gravity, at (a) \(6400 \mathrm{~m}\), and \((b) 6400 \mathrm{~km}\), above the Earth's surface.

Short Answer

Expert verified
At 6400 m, \( g' \approx 9.79 \text{ m/s}^2 \), and at 6400 km, \( g' = 2.4525 \text{ m/s}^2 \).

Step by step solution

01

Understand the Problem

We need to find the gravitational acceleration at 6400 meters and at 6400 kilometers above the Earth's surface. The gravitational acceleration decreases with altitude according to the formula: \[ g' = \frac{g_0}{(1 + \frac{h}{R})^2} \] where \( g_0 \) is the standard gravitational acceleration at Earth's surface, approximately 9.81 \( \text{m/s}^2 \), \( h \) is the height above the Earth's surface, and \( R \) is Earth's radius, about 6400 kilometers.
02

Convert Unit for Height

For part (a), the height \( h = 6400 \) meters, which is \( 6.4 \) km. For part (b), the height \( h = 6400 \) km is already given in kilometers, aligning with Earth's radius \( R = 6400 \) km.
03

Calculate Gravitational Acceleration for 6400 m

Substitute \( h = 6.4 \) km and \( R = 6400 \) km into the formula for \( g' \):\[ g' = \frac{9.81}{(1 + \frac{6.4}{6400})^2} \]Calculate the value inside the parentheses: \( 1 + \frac{6.4}{6400} = 1.001 \).Square the result: \( 1.001^2 = 1.002001 \).Compute \( g' \):\[ g' \approx \frac{9.81}{1.002001} \approx 9.79 \text{ m/s}^2 \].
04

Calculate Gravitational Acceleration for 6400 km

Substitute \( h = 6400 \) km into the formula for \( g' \):\[ g' = \frac{9.81}{(1 + \frac{6400}{6400})^2} \]\( 1 + \frac{6400}{6400} = 2 \) and \( 2^2 = 4 \).Compute \( g' \):\[ g' = \frac{9.81}{4} = 2.4525 \text{ m/s}^2 \].

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Key Concepts

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

Altitude Effect on Gravity
As you move away from Earth鈥檚 surface, the force of gravity weakens. This change in gravitational force is because gravity inversely depends on the distance from the Earth's center. At higher altitudes, like 6,400 meters (6.4 kilometers or km) and 6,400 km, the gravitational acceleration \(g'\) decreases from its standard value at the surface, which is about \(9.81\, ext{m/s}^2\).
  • At 6,400 meters above the surface, gravity decreases slightly because the altitude is small compared to Earth鈥檚 radius (6,400 km).
  • However, at 6,400 km (which is as far from the Earth's surface as the Earth's radius itself), gravity significantly decreases to about \(2.45 \, ext{m/s}^2\). This shows that the increase in altitude results in a more dramatic reduction in gravitational pull.
Therefore, understanding the altitude effect on gravity is crucial in fields like aerospace and geophysics, where precise calculations are necessary for navigation and satellite operations.
Calculation of Gravity
The calculation of gravitational acceleration at different altitudes uses the formula: \(g' = \frac{g_0}{(1 + \frac{h}{R})^2}\). This formula helps determine how gravity changes with altitude.
For example:
  • When calculating for an altitude of 6,400 meters (6.4 km), the formula becomes \(g' = \frac{9.81}{(1 + \frac{6.4}{6400})^2}\). After performing the calculations, \(g'\) is found to be approximately \(9.79\, ext{m/s}^2\).
  • At 6,400 km, the computation changes to \(g' = \frac{9.81}{4}\), which simplifies to \(2.45\, ext{m/s}^2\).
This demonstrates how gravity diminishes with increasing distance from Earth's surface. The gravitational acceleration \(g'\) is significantly impacted by both altitude and Earth's radius. These calculations are essential for understanding how gravity itself behaves across different heights, useful in areas like physics and engineering.
Earth's Radius Influence on Gravity
Earth鈥檚 radius plays a significant role in influencing gravitational acceleration. With a standard radius of about 6,400 km, Earth provides a frame of reference for gravitational calculations. The radius helps in understanding the gravity variations observed with different altitudes.
Here's how Earth's radius influences gravity:
  • In the formula \(g' = \frac{g_0}{(1 + \frac{h}{R})^2}\), \(R\) denotes the Earth's radius. A larger \(R\) would indicate a gentler decrease in gravitational force with altitude, while a smaller \(R\) would lead to more rapid decreases.
  • At higher altitudes, for example, 6,400 km from the Earth's surface (which is equal to Earth鈥檚 radius), gravity decreases markedly, as seen from the result \(g'\approx 2.45 \text{ m/s}^2\).
  • This shows that as one moves a distance equal to the Earth's radius away from the surface, gravitational acceleration is profoundly affected, decreasing to fractions of its surface value.
Therefore, Earth's radius is a fundamental component in calculating gravity and helps in visualizing the planet's influence on the gravitational forces experienced at various distances.

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Most popular questions from this chapter

Astronomers have observed an otherwise normal star, called S2, closely orbiting an extremely massive but small object at the center of the Milky Way Galaxy called SgrA. S2 moves in an elliptical orbit around SgrA with a period of \(15.2 \mathrm{yr}\) and an eccentricity \(e=0.87\) (Fig. \(6-16\) ). In \(2002, \mathrm{~S} 2\) reached its closest approach to SgrA, a distance of only \(123 \mathrm{AU} \quad\left(1 \mathrm{AU}=1.50 \times 10^{11} \mathrm{~m}\right.\) is the mean Earth-Sun distance). Determine the mass \(M\) of SgrA, the massive compact object (believed to be a supermassive black hole) at the center of our Galaxy. State \(M\) in \(\mathrm{kg}\) and in terms of the mass of our Sun.

(III) An inclined plane, fixed to the inside of an elevator, makes a \(32^{\circ}\) angle with the floor. A mass \(m\) slides on the plane without friction. What is its acceleration relative to the plane if the elevator \((a)\) accelerates upward at \(0.50 g\), (b) accelerates downward at \(0.50 g,(c)\) falls freely, and (d) moves upward at constant speed?

At the surface of a certain planet, the gravitational acceleration \(g\) has a magnitude of \(12.0 \mathrm{~m} / \mathrm{s}^{2} .\) A \(13.0-\mathrm{kg}\) brass ball is transported to this planet. What is ( \(a\) ) the mass of the brass ball on the Earth and on the planet, and (b) the weight of the brass ball on the Earth and on the planet?

(II) What will a spring scale read for the weight of a \(53-\mathrm{kg}\) woman in an elevator that moves \((a)\) upward with constant speed \(5.0 \mathrm{~m} / \mathrm{s},(b)\) downward with constant speed \(5.0 \mathrm{~m} / \mathrm{s}\) (c) upward with acceleration \(0.33 g,(d)\) downward with acceleration \(0.33 g\), and \((e)\) in free fall?

(II) (a) What is the gravitational field at the surface of the Earth due to the Sun? (b) Will this affect your weight significantly?
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