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Ceres, the largest asteroid in our solar system, is a spherical body with a mass 6000 times less than the earth's, and a radius which is 13 times smaller. If an astronaut who weighs \(400 \mathrm{~N}\) on earth is visiting the surface of Ceres, what is her weight?

Short Answer

Expert verified
The astronaut weighs approximately 11.27 N on Ceres.

Step by step solution

01

Understanding Gravitational Force

The gravitational force (weight) on an object is given by the formula: \( F = \frac{G M m}{r^2} \), where: \( F \) is the gravitational force, \( G \) is the gravitational constant, \( M \) is the mass of the celestial body, \( m \) is the mass of the object, and \( r \) is the radius of the celestial body.
02

Define Weight on Earth

For Earth, the weight of an object is \( F_{earth} = mg \), where \( m \) is the mass of the object and \( g \) is the acceleration due to gravity (approximately \( 9.8 \mathrm{~m/s^2} \)). Here, the astronaut's weight is given as \( 400 \mathrm{~N} \).
03

Relate Mass and Radius of Ceres to Earth

Ceres has a mass \( M_{ceres} = \frac{M_{earth}}{6000} \), and a radius \( r_{ceres} = \frac{r_{earth}}{13} \).
04

Calculate Weight on Ceres

Using the formula \( F = \frac{G M m}{r^2} \), the weight on Ceres can be written as : \[ F_{ceres} = \frac{G \left( \frac{M_{earth}}{6000} \right) m}{\left( \frac{r_{earth}}{13} \right)^2} \] Simplifying this gives: \[ F_{ceres} = \frac{G M_{earth} m}{6000} \times \frac{169}{r_{earth}^2} \] Simplifying further using \( F_{earth} = \frac{G M_{earth} m}{r_{earth}^2} = 400 \mathrm{~N}\): \[ F_{ceres} = 400 \frac{169}{6000} \approx 11.27 \mathrm{~N} \]
05

Interpretation of Results

The calculation shows that her weight on Ceres is approximately \( 11.27 \mathrm{~N} \). This is significantly lower than her weight on Earth due to Ceres' much smaller mass and size.

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Key Concepts

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

Ceres
Ceres is an intriguing and fascinating celestial body located within our solar system. It's situated in the asteroid belt between the orbits of Mars and Jupiter. Known as the largest asteroid, Ceres accounts for a significant portion of the mass of the asteroid belt. It is unique because it is also classified as a dwarf planet due to its spherical shape.
Unlike a typical asteroid, Ceres' spherical form gives it planet-like characteristics. Despite its considerable size for an asteroid, Ceres is about 6000 times less massive than Earth, and its radius is approximately 13 times smaller than that of Earth.
Because of its reduced mass, the gravitational pull on Ceres is much weaker than that on Earth. This is why an object or person would weigh significantly less on Ceres than on Earth, making it an intriguing subject for gravitational studies.
Weight Calculation
Weight is the force exerted on a mass due to gravity. It is crucial to understand that weight is different from mass. Weight depends on the gravitational pull exerted by the celestial body you are on.
To calculate weight, we use the formula:
  • \( F = \frac{G M m}{r^2} \)
  • Where:
    • \( F \) is the gravitational force or weight,
    • \( G \) is the gravitational constant,
    • \( M \) is the mass of the celestial body,
    • \( m \) is the mass of the object,
    • \( r \) is the radius of the celestial body.
When calculating weight on Earth, we usually simplify this into \( F = mg \) given Earth's known gravity. However, for celestial objects like Ceres, you need to account for the differing mass and radius to determine the weight there.
Gravitational Constant
The gravitational constant, represented by \( G \), is a fundamental constant in physics. It plays a crucial role in the calculation of gravitational forces in the universe.
Its value is approximately \(6.674 \times 10^{-11} \, \mathrm{m^3 \, kg^{-1} \, s^{-2}}\).
Every interaction involving gravity uses \( G \) to relate the masses of two bodies and the distance between them, hence allowing us to calculate the gravitational force accurately.
In our specific scenario, \( G \) is used to determine the gravitational force between an object or person and Ceres. By incorporating \( G \) with the mass and radius of Ceres along with the object's mass, we successfully find the object's weight on Ceres.
Acceleration Due to Gravity
Acceleration due to gravity, represented by \( g \), is the rate at which an object accelerates when gravitational force is applied. On Earth, this is approximately \( 9.8 \, \mathrm{m/s^2} \).
For other celestial bodies like Ceres, \( g \) differs significantly due to variations in mass and radius. The formula to calculate \( g \) on any celestial body is given by:
  • \( g_{body} = \frac{G M_{body}}{r_{body}^2} \)
  • Where:
    • \( G \) is the gravitational constant,
    • \( M_{body} \) is the mass of the celestial body,
    • \( r_{body} \) is the radius of the celestial body.
On Ceres, this calculated \( g \) is much lower than Earth's, leading to significantly reduced weights. Understanding this concept is vital for exploring how gravity affects everything from falling objects to the weight one would experience on different planets or asteroids like Ceres.

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Most popular questions from this chapter

On an airless body such as the moon, there is no atmospheric friction, so it should be possible for a satellite to orbit at a very low altitude, just high enough to keep from hitting the mountains. (a) Suppose that such a body is a smooth sphere of uniform density \(\rho\) and radius \(r .\) Find the velocity required for a ground-skimming orbit. (b) A typical asteroid has a density of about \(2 \mathrm{~g} / \mathrm{cm}^{3}\), i.e., twice that of water. (This is a lot lower than the density of the earth's crust, probably indicating that the low gravity is not enough to compact the material very tightly, leaving lots of empty space inside.) Suppose that it is possible for an astronaut in a spacesuit to jump at \(2 \mathrm{~m} / \mathrm{s}\). Find the radius of the largest asteroid on which it would be possible to jump into a ground-skimming orbit.

How high above the Earth's surface must a rocket be in order to have \(1 / 100\) the weight it would have at the surface? Express your answer in units of the radius of the Earth.

Prove, based on Newton's laws of motion and Newton's law of gravity, that all falling objects have the same acceleration if they are dropped at the same location on the earth and if other forces such as friction are unimportant. Do not just say, " \(g=9.8 \mathrm{~m} / \mathrm{s}^{2}-\) it's constant." You are supposed to be proving that \(g\) should be the same number for all objects.

(a) Suppose a rotating spherical body such as a planet has a radius \(r\) and a uniform density \(\rho\), and the time required for one rotation is \(T\). At the surface of the planet, the apparent acceleration of a falling object is reduced by the acceleration of the ground out from under it. Derive an equation for the apparent acceleration of gravity, \(g\), at the equator in terms of \(r, \rho, T\), and \(G .\) (b) Applying your equation from a, by what fraction is your apparent weight reduced at the equator compared to the poles, due to the Earth's rotation? (c) Using your equation from a, derive an equation giving the value of \(T\) for which the apparent acceleration of gravity becomes zero, i.e., objects can spontaneously drift off the surface of the planet. Show that \(T\) only depends on \(\rho\), and not on \(r\). (d) Applying your equation from \(\mathrm{c}\), how long would a day have to be in order to reduce the apparent weight of objects at the equator of the Earth to zero? [Answer: \(1.4\) hours \(]\) (e) Astronomers have discovered objects they called pulsars, which emit bursts of radiation at regular intervals of less than a second. If a pulsar is to be interpreted as a rotating sphere beaming out a natural "searchlight" that sweeps past the earth with each rotation, use your equation from \(\mathrm{c}\) to show that its density would have to be much greater than that of ordinary matter. (f) Astrophysicists predicted decades ago that certain stars that used up their sources of energy could collapse, forming a ball of neutrons with the fantastic density of \(\sim 10^{17} \mathrm{~kg} / \mathrm{m}^{3} .\) If this is what pulsars really are, use your equation from \(c\) to explain why no pulsar has ever been observed that flashes with a period of less than \(1 \mathrm{~ms}\) or so.

The International Space Station orbits at an average altitude of about \(370 \mathrm{~km}\) above sea level. Compute the value of \(g\) at that altitude.

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