/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 15 Calculate the earth's gravity fo... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Calculate the earth's gravity force on a 75 -kg astronaut who is repairing the Hubble Space Telescope 600 \(\mathrm{km}\) above the earth's surface, and then compare this value with his weight at the earth's surface. In view of your result, explain why we say astronauts are weightless when they orbit the earth in a satellite such as a space shuttle. Is it because the gravitational pull of the earth is negligibly small?

Short Answer

Expert verified
The gravitational force at 600 km is less but not negligible. Astronauts feel weightless because they are in free fall, not because gravity is negligible.

Step by step solution

01

Identify Given Values and Constants

The astronaut's mass (\( m \)) is 75 kg. The distance above Earth's surface is 600 km, which is 600,000 meters. The radius of Earth (\( R \)) is approximately 6,371,000 meters. The gravitational constant (\( G \)) is \( 6.674 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \). The Earth's mass (\( M \)) is \( 5.972 \times 10^{24} \, \text{kg} \). With these, we can calculate the gravitational force.
02

Calculate Gravitational Force at 600 km Above Earth's Surface

First, find the total distance from the center of Earth to the astronaut, which is \( R + 600,000 \) meters. Then use the formula for gravitational force: \[ F = \frac{G M m}{(R + 600,000)^2} \].Substitute the given values:\[ F = \frac{6.674 \times 10^{-11} \times 5.972 \times 10^{24} \times 75}{(6,371,000 + 600,000)^2} \].Calculate this to find \( F \).
03

Calculate Weight on Earth's Surface

The weight of the astronaut on Earth's surface is calculated using: \[ W = mg \], where the acceleration due to gravity on the Earth's surface \( g \) is \( 9.81 \, \text{m/s}^2 \). This gives:\[ W = 75 \times 9.81 \].Compute this to find the weight on the surface.
04

Compare the Two Forces

Calculate the ratio of gravitational force at 600 km above the surface to the weight on Earth's surface:\[ \frac{F}{W} \].Evaluate this ratio to understand the relative strength of gravity at 600 km compared to Earth's surface gravity.
05

Explain the Concept of Weightlessness

Astronauts feel weightless not because the gravitational pull is negligibly small but because they are in free fall, orbiting the Earth. In orbit, they are constantly falling towards Earth but also moving forward, creating a sensation of weightlessness.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Astronaut Weight
When we talk about an astronaut's weight, we're referring to the force exerted by gravity on their mass. On Earth, this force is straightforward: it's simply their mass multiplied by the acceleration due to Earth's gravity, or
  • Weight = mass × gravity (which is about 9.81 m/s²).
However, when an astronaut is in space, especially at a height of 600 km above the Earth's surface while repairing something like the Hubble Space Telescope, calculating this becomes a bit more complex.

At this altitude, the force of gravity is still there but is weaker than on the ground.
The astronaut's "weight" as felt by them isn't the same as it would be on Earth.
This might seem counterintuitive since we often assume they are weightless in space, but we'll get into why this is the case later.
Earth's Gravity
Earth's gravity is the force that keeps everything anchored to our planet. It pulls objects towards the Earth's center. The strength of this force diminishes with distance from the Earth's surface.
  • Near the surface, gravity is strong enough to keep everything, from people to oceans, firmly in place.
When you are still relatively close to Earth, like a few hundred kilometers up, gravity still has a strong hold.
The gravitational force can be calculated using the equation: \[ F = \frac{G M m}{R^2} \]where
  • \( G \) is the universal gravitational constant, approximately \( 6.674 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \),
  • \( M \) is Earth's mass,
  • \( m \) is the object's mass, and
  • \( R \) is the distance from the center of the Earth.
Therefore, even at 600 km above the Earth's surface, an astronaut is still under the influence of gravity, albeit slightly less than at ground level.
Orbital Mechanics
Orbital mechanics are the rules governing movements of objects in space under the influence of gravitational forces. When an astronaut is orbiting Earth, they are subject to these principles. Imagine throwing a ball so fast that rather than falling back to the ground, it keeps missing the Earth, following a curved path around it.
  • This is essentially what happens in orbit – the astronaut is in constant free fall towards Earth, but their forward motion keeps them in a continuous path around the planet.
In this state, the astronaut and their spacecraft are moving at such speeds that their downward motion due to gravity is perfectly matched by their forward motion.

This combination keeps them floating along a stable path, like dancing with destiny, around the Earth.
Weightlessness in Space
Despite gravity still acting on astronauts, they experience a sensation of weightlessness. This isn't because the gravitational force is negligible. Instead, it's due to their condition of "free fall", which is the same state achieved when skydiving briefly before a parachute deploys.
  • As both the astronaut and their ship are falling, everything inside moves at the same rate, effectively "floating" relative to one another.
This is why astronauts appear to be weightless while working in space despite Earth's gravity still pulling on them. They are in a state of continuous free fall, causing them to feel as though they carry no weight at all.

Weightlessness thus is more accurately a term for the synchronized free-fall of an astronaut and their environment, rather than an absence of gravitational force.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The earth does not have a uniform density; it is most dense at its center and least dense at its surface. An approximation of its density is \(\rho(r)=A-B r,\) where \(A=12,700 \mathrm{kg} / \mathrm{m}^{3}\) and \(B=\) \(1.50 \times 10^{-3} \mathrm{kg} / \mathrm{m}^{4}\) . Use \(R=6.37 \times 10^{6} \mathrm{m}\) for the radius of the earth approximated as a sphere. (a) Geological evidence indicates that the densities are \(13,100 \mathrm{kg} / \mathrm{m}^{3}\) and \(2,400 \mathrm{kg} / \mathrm{m}^{3}\) at the earth's center and surface, respectively. What values does the linear approximation model give for the densities at these two locations? (b) Imagine dividing the earth into concentric, spherical shells. Each shell has radius \(r\) , thickness \(d r\) , volume \(d V=4 \pi r^{2} d r,\) and mass \(d m=\rho(r) d V .\) By integrating from \(r=0\) to \(r=R,\) show that the mass of the earth in this model is \(M=\frac{4}{3} \pi R^{3}\left(A-\frac{3}{4} B R\right)\) (c) Show that the given values of \(A\) and \(B\) give the correct mass of the earth to within 0.4\(\%\) (d) We saw in Section 13.6 that a uniform spherical shell gives no contribution to \(g\) inside it. Show that \(g(r)=\frac{4}{3} \pi G r\left(A-\frac{3}{4} B r\right)\) inside the earth in this model. (e) Verify that the expression of part (d) gives \(g=0\) at the center of the earth and \(g=9.85 \mathrm{m} / \mathrm{s}^{2}\) at the surface. (f) Show that in this model \(g\) does not decrease uniformly with depth but rather has a maximum of \(4 \pi G A^{2} / 9 B=10.01 \mathrm{m} / \mathrm{s}^{2}\) at \(r=2 A / 3 B=5640 \mathrm{km} .\)

Two satellites are in circular orbits around a planet that has radius \(9.00 \times 10^{6} \mathrm{m}\) . One satellite has mass 68.0 \(\mathrm{kg}\) , orbital radius \(5.00 \times 10^{7} \mathrm{m},\) and orbital speed 4800 \(\mathrm{m} / \mathrm{s} .\) The second satellite has mass 84.0 \(\mathrm{kg}\) and orbital radius \(3.00 \times 10^{7} \mathrm{m} .\) What is the orbital speed of this second satellite?

Comets travel around the sun in elliptical orbits with large eccentricities. If a comet has speed \(2.0 \times 10^{4} \mathrm{m} / \mathrm{s}\) when at a distance of \(2.5 \times 10^{11} \mathrm{m}\) from the center of the sun, what is its speed when at a distance of \(5.0 \times 10^{10} \mathrm{m} ?\)

Binary Star-Equal Masses. Two identical stars with mass \(M\) orbit around their center of mass. Each orbit is circular and has radius \(R,\) so that the two stars are always on opposite sides of the circle. (a) Find the gravitational force of one star on the other. (b) Find the orbital speed of each star and the period of the orbit. (c) How much energy would be required to separate the two stars to infinity?

Cavendish Experiment. In the Cavendish balance apparatus shown in Fig. \(13.4,\) suppose that \(m_{1}=1.10 \mathrm{kg}, m_{2}=\) \(25.0 \mathrm{kg},\) and the rod connecting the \(m_{1}\) pairs is 30.0 \(\mathrm{cm}\) long. If, in each pair, \(m_{1}\) and \(m_{2}\) are 12.0 \(\mathrm{cm}\) apart center to center, find (a) the net force and (b) the net torque (about the rotation axis) on the rotating part of the apparatus. (c) Does it seem that the torque in part (b) would be enough to easily rotate the rod? Suggest some ways to improve the sensitivity of this experiment.

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.