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Chapter 13: Q. 20 (page 354)

Nothing can escape the event horizon of a black hole, not even light. You can think of the event horizon as being the distance from a black hole at which the escape speed is the speed of light, 3.00*108m/s,making all escape impossible. What is the radius of the event horizon for a black hole with a mass 5.0times the mass of the sun? This distance is called the Schwarzschild radius.

Short Answer

Expert verified

The radius of the event horizon of the black hole is14.74km.

Step by step solution

01

Given information

Mass of black hole = 5times of the mass of the sun, escape velocity =3脳108m/s.

02

Calculation

We know, escape velocity : v=2G脳MR.

Substituting the values of G, M, v in the above equation, and solving for R we get :

localid="1648487065595" 3脳108m/s=2(6.67脳10-11Nm2/kg2)(5脳1.989脳1030kg)R9脳1016m/s=2(6.67脳10-11Nm2/kg2)(5脳1.989脳1030kg)RR=2(6.67脳10-11Nm2/kg2)(5脳1.989脳1030kg)(9脳1016m/s)R=14740.7mR=14.740km.

03

Final answer

The radius of the event horizon for the black hole is 14.74km.

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Let鈥檚 look in more detail at how a satellite is moved from one circular orbit to another. FIGURE CP13.71shows two circular orbits, of radii localid="1651418485730" r1and localid="1651418489556" r2, and an elliptical orbit that connects them. Points 1and 2are at the ends of the semimajor axis of the ellipse.

a. A satellite moving along the elliptical orbit has to satisfy two conservation laws. Use these two laws to prove that the velocities at points localid="1651418503699" 1and localid="1651418499267" 2are localid="1651418492993" v1鈥�=2GMr2/r1r1+r2and localid="1651418509687" v2鈥�=2GMr1/r2r1+r2The prime indicates that these are the velocities on the elliptical orbit. Both reduce to Equation 13.22if localid="1651418513535" r1=r2=r.

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