/*! 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} Q47P The Sun, which is  2.2×1020 ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

The Sun, which is2.2×1020mfrom the center of the Milky Way galaxy, revolves around that center once every 2.5×108years. Assuming each star in the Galaxy has a mass equal to the Sun’s mass of 2.0×1030kg, the stars are distributed uniformly in a sphere about the galactic center, and the Sun is at the edge of that sphere, estimate the number of stars in the Galaxy.

Short Answer

Expert verified

The number of stars in the galaxy are 5.1×1010.

Step by step solution

01

Step 1: Given

The sun is 2.2×1020mfar from the center of the Milky Way galaxy

The sun revolves around that center once every 2.5×108years

Each star in the galaxy has a mass equal to the sun’s mass of 2.0×1030

02

Determining the concept

Using the formula of gravitational force, centripetal force and period, find the acceleration of the sun’s motion around the galactic center. Then, using Newton’s second law, find the number of the stars.

Formulae are as follow:

Fg=GMmR2

FC=Mv2R

V=2Ï€¸éT

where, M, m are masses, R is radius, T is time, G is gravitational constant, v, V are velocities and F is corresponding force.

03

Determining the number of stars in the galaxy

Consider the total mass in the galaxy as m=NM ,

where, N is the number of stars in the galaxy and M is the mass of the sun.

Now,

Fg=GMmR2

Therefore,

Fg=GMNMR2

Fg=GNM2R2

The centripetal force on the sun is pointing towards the galactic center,

FC=Mv2R=Fg

Mv2R=GNM2R2

IfTis the period of the sun’s motion around the galactic center then,

T=2Ï€¸éV

V=2Ï€¸éT

But,

a=V2R

a=4Ï€2RT2

Therefore, according to Newton’s second law,

GNM2R2=4Ï€2MRT2

N=4Ï€2R3GT2M

As,

°Õ â¶Ä‰= â¶Ä‰2.5×108years=7.88×1015 s

N=4Ï€2(2.2×1020″¾)3(6.67×10-11″¾3/s2â‹…kg)(7.88×1015 s)2(2.0×1030 kg)

N=5.1×1010

Hence, the number of starts in the galaxy are 5.1×1010.

Therefore, using the formula of gravitational force, centripetal force and period, the acceleration of the sun around the galactic center can be found. Using Newton’s second law, the number of stars in the galaxy can be found.

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!

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

Two small spaceships, each with massm=2000kg, are in the circular Earth orbit of the figure, at an altitudehof400km.Igor, the commander of one of the ships, arrives at any fixed point in the orbit90sahead of Picard, the commander of the other ship. What are the (a) periodT0and (b) speedv0of the ships? At point P in the figure, Picard fires an instantaneous burst in the forward direction, reducing his ship’s speed by1.00%.after this burst; he follows the elliptical orbit shown dashed in the figure. What are the(c) kinetic energy and (d) potential energy of his ship immediately after the burst? In Picard’s new elliptical orbit, what are (e) the total energyE,(f) the semi major axisrole="math" localid="1661171269628" a, and(g) the orbital periodT?(h) How much earlier than Igor will Picard return toP?

One way to attack a satellite in Earth orbit is to launch a swarm of pellets in the same orbit as the satellite but in the opposite direction. Suppose a satellite in a circular orbit 500 kmabove Earth’s surface collides with a pellet having mass 4.0g.

(a) What is the kinetic energy of the pellet in the reference frame of the satellite just before the collision?

b) What is the ratio of this kinetic energy to the kinetic energy of a 4.0gbullet from a modern army rifle with a muzzle speed of 950m/s?

In Fig. 13-50, two satellites, A and B, both of mass m=125kg , move in the same circular orbit of radius r=7.87×106maround Earth but in opposite senses of rotation and therefore on a collision course.

(a) Find the total mechanical energy role="math" localid="1661161625366" EA+EBof thetwosatellites+Earth system before the collision.

(b) If the collision is completely inelastic so that the wreckage remains as one piece of tangled material ( mass=2m), find the total mechanical energy immediately after the collision.

(c) Just after the collision, is the wreckage falling directly toward Earth’s center or orbiting around Earth?

Figure 13-22 shows three arrangements of the same identical particles, with three of them placed on a circle of radius 0.20mand the fourth one placed at the center of the circle. (a) Rank the arrangements according to the magnitude of the net gravitational force on the central particle due to the other three particles, greatest first. (b) Rank them according to the gravitational potential energy of the four-particle system, least negative first.

One dimension.In the figure, two point particles are fixed on anxaxis separated by distanced. ParticleAhas massmAM and particle Bhas mass3.00mA. A third particle C, of mass750mA, is to be placed on the xaxis and near particles Aand B. In terms of distance d, at what xcoordinate should Cbe placed so that the net gravitational force on particle Afrom particles Band Cis zero?

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.