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Chapter 12: Problem 82

The orbit of the planet Venus is nearly circular with an orbital velocity of \(126.5 \times 10^{3} \mathrm{km} / \mathrm{h}\). Knowing that the mean distance from the center of the sun to the center of Venus is \(108 \times 10^{6} \mathrm{km}\) and that the radius of the sun is \(695.5 \times 10^{3} \mathrm{km}\), determine (a) the mass of the sun, \((b)\) the acceleration of gravity at the surface of the sun.

Short Answer

Expert verified
(a) Mass of the sun: \(1.989 \times 10^{30} \mathrm{kg}\). (b) Acceleration of gravity at the sun's surface: \(274 \mathrm{m/s}^2\).

Step by step solution

01

Determine Centripetal Force on Venus

To find the mass of the sun, we need the centripetal force acting on Venus due to its orbital motion. The centripetal force \( F_c \) is given by \( F_c = \frac{mv^2}{R} \), where \( m \) is the mass of Venus, \( v \) is the orbital speed, and \( R \) is the radius of the orbit (distance from the center of the sun to Venus). We will use the gravitational force as it provides the centripetal force, \( F_c = G \frac{Mm}{R^2} \). Equating both expressions for force and solving for \( M \), the mass of the sun.
02

Equate Gravitational and Centripetal Forces

The equation \( G \frac{Mm}{R^2} = \frac{mv^2}{R} \) allows us to cancel out the mass of Venus (\( m \)) and one power of the radius (\( R \)), simplifying to \( M = \frac{Rv^2}{G} \). We substitute \( R = 108 \times 10^{6} \mathrm{km} = 108 \times 10^{9} \mathrm{m} \) and \( v = 126.5 \times 10^{3} \mathrm{km/h} = 126.5 \times 10^{3} \times \frac{1000}{3600} = 3.514 \times 10^{4} \mathrm{m/s} \).
03

Calculate Mass of the Sun

Use the gravitational constant \( G = 6.674 \times 10^{-11} \mathrm{N} \cdot \mathrm{m}^2/\mathrm{kg}^2 \) to find \( M = \frac{108 \times 10^{9} \times (3.514 \times 10^{4})^2}{6.674 \times 10^{-11}}. \) Solving gives \( M \approx 1.989 \times 10^{30} \mathrm{kg} \).
04

Determine Acceleration of Gravity at the Sun's Surface

The acceleration due to gravity \( g \) at the surface of the sun is given by \( g = \frac{GM}{R_s^2} \), where \( R_s \) is the radius of the sun. Use \( R_s = 695.5 \times 10^{3} \mathrm{km} = 695.5 \times 10^{6} \mathrm{m} \) and \( M = 1.989 \times 10^{30} \mathrm{kg} \).
05

Calculate Acceleration of Gravity

Substitute the values into the equation \( g = \frac{6.674 \times 10^{-11} \times 1.989 \times 10^{30}}{(695.5 \times 10^{6})^2} \) to find \( g \approx 274 \mathrm{m/s}^2.\)

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Key Concepts

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

Centripetal Force
In orbital mechanics, the concept of centripetal force is fundamental to understanding the motion of celestial bodies, like Venus around the Sun. Centripetal force is necessary to keep an object moving in a circular path. It acts towards the center of the circle and ensures that the object doesn't fly off in a straight line.
For Venus, orbiting nearly circularly around the Sun, its centripetal force comes from the gravitational pull of the Sun. This force keeps Venus in a fixed path, preventing it from veering into space.
Here's how we calculate it:
  • The centripetal force (\( F_c \)) is described by the equation: \( F_c = \frac{mv^2}{R} \), where:
    • \( m \) is the mass of Venus,
    • \( v \) is its velocity,
    • \( R \) is the radius of its orbit.
  • However, Venus's mass cancels out in these calculations, focusing solely on its velocity and distance from the Sun.
The gravitational force exerted by the Sun provides this centripetal force, allowing us to equate these forces and solve for related quantities, such as the mass of the Sun.
Gravitational Force
Gravitational force is the attraction between two masses. It's the same force that keeps us on Earth and planets in orbit. Governed by Newton's law of gravitation, this force depends on the masses involved and the distance between them.
The formula for calculating gravitational force (\( F_g \)) is:
  • \( F_g = G \frac{Mm}{R^2} \)
    • \( G \) is the gravitational constant \( (6.674 \times 10^{-11} \text{ N} \cdot \text{m}^2/\text{kg}^2) \).
    • \( M \) and \( m \) are the two masses (like the Sun and Venus),
    • \( R \) is the distance between their centers.
This force plays a crucial role in keeping celestial bodies in orbit. In the case of Venus, the gravitational pull from the Sun acts as the centripetal force needed to maintain its nearly circular orbit. The elegant part of this scenario is how we can use these forces to deduce key properties like the mass of the Sun, giving us insights into the mechanics of our solar system.
Acceleration of Gravity
The acceleration of gravity (\( g \)) refers to the rate at which objects accelerate towards a massive body due to its gravitational pull. On Earth, this acceleration is approximated to \( 9.81 \text{ m/s}^2 \), meaning any object will gain speed as it falls towards the planet. However, on the Sun, this acceleration is far more intense.
To find the acceleration of gravity on the Sun's surface, we employ the formula:
  • \( g = \frac{GM}{R_s^2} \)
    • \( R_s \) is the radius of the Sun. For this exercise, \( R_s = 695.5 \times 10^6 \text{ m} \).
By computing with the mass of the Sun and the gravitational constant, we find that the acceleration of gravity at the Sun's surface is approximately \( 274 \text{ m/s}^2 \).
This value illustrates just how strong the Sun's gravitational pull is compared to Earth's. Such a high gravitational acceleration indicates that an object would require immense energy to overcome this pull and ascend from the Sun's surface.

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Most popular questions from this chapter

A \(0.5-\mathrm{kg}\) block \(B\) slides without friction inside a slot cut in arm \(O A\) that rotates in a vertical plane. The rod has a constant angular acceleration \(\dot{\theta}=10 \mathrm{rad} / \mathrm{s}^{2} .\) Knowing that when \(\theta=45^{\circ}\) and \(r=0.8 \mathrm{m}\) the velocity of the block is zero, determine at this instant, (a) the force exerted on the block by the arm, (b) the relative acceleration of the block with respect to the arm.

Block \(A\) has a mass of \(10 \mathrm{kg}\), and blocks \(B\) and \(C\) have masses of \(5 \mathrm{kg}\) each. Knowing that the blocks are initially at rest and that \(B\) moves through \(3 \mathrm{m}\) in \(2 \mathrm{s}\), determine (a) the magnitude of the force \(\mathrm{P},(b)\) the tension in the cord \(A D .\) Neglect the masses of the pulleys and axle friction.

A constant force \(\mathbf{P}\) is applied to a piston and rod of total mass \(m\) to make them move in a cylinder filled with oil. As the piston moves, the oil is forced through orifices in the piston and exerts on the piston a force of magnitude \(k v\) in a direction opposite to the motion of the piston. Knowing that the piston starts from rest at \(t=0\) and \(x=0,\) show that the equation relating \(x, v,\) and \(t,\) where \(x\) is the distance traveled by the piston and \(v\) is the speed of the piston, is linear in each of these variables.

A 1 -kg collar can slide on a horizontal rod that is free to rotate about a vertical shaft. The collar is initially held at \(A\) by a cord attached to the shaft. A spring of constant \(30 \mathrm{N} / \mathrm{m}\) is attached to the collar and to the shaft and is undeformed when the collar is at \(A .\) As the rod rotates at the rate \(\dot{\theta}=16 \mathrm{rad} / \mathrm{s}\), the cord is cut and the collar moves out along the rod. Neglecting friction and the mass of the rod, determine (a) the radial and transverse components of the acceleration of the collar at \(A,(b)\) the acceleration of the collar relative to the rod at \(A,(c)\) the transverse component of the velocity of the collar at \(B .\)

A 0.2 -lb model rocket is launched vertically from rest at time \(t=0\) with a constant thrust of 2 lb for one second and no thrust for \(t>1\) s. Neglecting air resistance and the decrease in mass of the rocket, determine ( \(a\) ) the maximum height \(h\) reached by the rocket, \((b)\) the time required to reach this maximum height.
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