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Chapter 13: Problem 12

The mass of Venus is 81.5\(\%\) that of the earth, and its radius is 94.9\(\%\) that of the earth. (a) Compute the acceleration due to gravity on the surface of Venus from these data. (b) If a rock weighs 75.0 \(\mathrm{N}\) on earth, what would it weigh at the surface of Venus?

Short Answer

Expert verified
Venus's surface gravity is approximately 8.88 m/s虏, and the rock would weigh about 67.92 N.

Step by step solution

01

Understand the formula for gravitational acceleration

The formula for gravitational acceleration on a planet's surface is given by \( g = \frac{GM}{R^2} \), where \( G \) is the gravitational constant, \( M \) is the mass of the planet, and \( R \) is the radius of the planet.
02

Compare Venus and Earth's parameters

The mass of Venus is 81.5\(\%\) of Earth's mass and the radius of Venus is 94.9\(\%\) of Earth's radius. Let \( M_e \) represent Earth's mass and \( R_e \) represent Earth's radius. So, the mass of Venus \( M_v = 0.815M_e \) and the radius of Venus \( R_v = 0.949R_e \).
03

Express Venus's gravitational acceleration using Earth's terms

The gravitational acceleration on Venus's surface can be expressed as \( g_{v} = \frac{G(0.815M_e)}{(0.949R_e)^2} \). To make this comparable to Earth's gravity, \( g = \frac{GM_e}{R_e^2} \), multiply out the terms: \( g_{v} = 0.815 \times (\frac{1}{0.949^2}) \times g_e \).
04

Calculate Venus's gravity as a fraction of Earth's gravity

Compute the value of \( 0.815 \times (\frac{1}{0.949^2}) \) which approximates to \( 0.905 \). Hence, \( g_{v} \approx 0.905g_e \). If Earth's gravity \( g_e \approx 9.81 \, \text{m/s}^2 \), then \( g_{v} \approx 0.905 \times 9.81 \, \text{m/s}^2 \approx 8.88 \, \text{m/s}^2 \).
05

Use gravitational force equation to find the weight on Venus

Weight is the product of mass and gravitational acceleration, \( W = mg \). If a rock weighs 75.0 N on Earth, its mass \( m = \frac{75}{9.81} \). On Venus, its weight \( W_v = m \times g_{v} = \frac{75}{9.81} \times 8.88 \).
06

Compute the rock's weight on Venus

Calculate the weight of the rock on Venus: \( W_v = \left( \frac{75}{9.81} \right) \times 8.88 \approx 67.92 \, \text{N} \).

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Key Concepts

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

Mass of Venus
The mass of Venus is a crucial factor in understanding its gravitational pull. Venus is often compared to Earth because of its size and composition. However, its mass is 81.5% that of Earth. To put this in perspective, if Earth's mass is represented as 1, Venus's mass is 0.815.
This difference in mass affects how objects experience gravity on Venus compared to Earth. Understanding this helps us calculate gravitational acceleration more accurately for any object on Venus.
Radius of Venus
The radius of a planet is an important part of determining its surface gravity. Venus's radius is 94.9% that of Earth's. This means Venus is slightly smaller in size compared to Earth.
Just like mass, the radius closely relates to gravitational calculations. Specifically, when calculating gravity on a planet鈥檚 surface, we use the planet's radius in the formula. The radius helps us understand the distance from the planet's core to its surface, impacting how strongly gravity acts on objects there.
Weight Calculation
Calculating weight on another planet requires understanding the planet's gravitational pull. Weight, scientifically, is the force of gravity acting on an object's mass, given by the equation:
  • Weight (W) = mass (m)
  • Gravitational acceleration (g)
If we have an object weighing 75 N on Earth, we understand its mass using Earth's gravity (approximately 9.81 m/s虏).
When calculating the same object鈥檚 weight on Venus, we use its gravitational acceleration, which we find to be roughly 8.88 m/s虏. This adjustment shows why the weight of objects changes from planet to planet.
Gravitational Force Equation
The gravitational force equation is fundamental to understanding how planets attract objects. This formula for gravitational acceleration is used to discover how strong gravity is on a planet's surface.
Given by \[ g = \frac{GM}{R^2} \]
  • 'G' is the gravitational constant, a universal number that applies to any celestial body.
  • 'M' represents the mass of the planet.
  • 'R' is the planet's radius.
This equation allows us to calculate gravity by considering both mass and radius. It's particularly helpful in comparing gravity between different planets, as we've done with Venus and Earth in this exercise.

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Most popular questions from this chapter

The point masses \(m\) and 2\(m\) lie along the \(x\) -axis, with \(m\) at the origin and 2\(m\) at \(x=L .\) A third point mass \(M\) is moved along the \(x\) -axis. (a) At what point is the net gravitational force on \(M\) due to the other two masses equal to zero? (b) Sketch the \(x\) -component of the net force on \(M\) due to \(m\) and \(2 m,\) taking quantities to the right as positive. Include the regions \(x < 0,0< x < L,\) and \(x>L .\) Be especially careful to show the behavior of the graph on either side of \(x=0\) and \(x=L\) .

A uniform sphere with mass 60.0 \(\mathrm{kg}\) is held with its center at the origin, and a second uniform sphere with mass 80.0 \(\mathrm{kg}\) is held with its center at the point \(x=0, y=3.00 \mathrm{m} .\) (a) What are the magnitude and direction of the net gravitational force due to these objects on a third uniform sphere with mass 0.500 \(\mathrm{kg}\) placed at the point \(x=4.00 \mathrm{m}, y=0 ?\) (b) Where, other than infinitely far away, could the third sphere be placed such that the net gravitational force acting on it from the other two spheres is equal to zero?

A uniform wire with mass \(M\) and length \(L\) is bent into a semicircle. Find the magnitude and direction of the gravitational force this wire exerts on a point with mass \(m\) placed at the center of curvature of the semicircle.

At what distance above the surface of the earth is the acceleration due to the earth's gravity 0.980 \(\mathrm{m} / \mathrm{s}^{2}\) if the acceleration due to gravity at the surface has magnitude 9.80 \(\mathrm{m} / \mathrm{s}^{2}\) ?

In March \(2006,\) two small satellites were discovered orbiting Pluto, one at a distance of \(48,000 \mathrm{km}\) and the other at \(64,000 \mathrm{km} .\) Pluto already was known to have a large satellite Charon, orbiting at \(19,600 \mathrm{km}\) with an orbital period of 6.39 days. Assuming that the satellites do not affect each other, find the orbital periods of the two small satellites without using the mass of Pluto.
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