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Planet Z is 10,000kmin diameter. The free-fall acceleration on Planet Z is role="math" localid="1648089747827" 8.0m/s2.

(a) What is the mass of Planet Z?

(b) What is the free-fall acceleration10,000kmabove Planet Z’s north pole?

Short Answer

Expert verified

(a) The mass of planet Z is1.25×1025kg.

(b) The free-fall acceleration 10000kmabove Planet Z’s north pole is 2m/s2.

Step by step solution

01

Given information (a)

Diameter of the planet = 10000km, free-fall acceleration = 8m/s2.

02

Calculation (a)

The formula for free-fall acceleration is given by :g=GMR2.

Substituting the given values, in equation :

role="math" localid="1648091089811" ⇒8.0m/s2=6.67×10-11N·m2/kg2M1.0×107m2.⇒M=8.0m/s21.0×107m26.67×10-11N·m2/kg2=1.2×1025kg.
03

Final answer (a)

The mass of planet Z is 1.25×1025kg.

04

Given information (b)

Height from the north pole h=10000km.

05

Calculation (b)

The formula for free-fall acceleration at a height h is given by :g'=GM(R+h)2.

Height given in this case is h=R, hence h=Ris substituted to the above equation :

g'=GM(R+R)2=GM4R2=14GMR2=14g=148.0m/s2=2.0m/s2.

06

Final answer (b)

The free-fall acceleration 10000kmabove Planet Z’s north pole is2m/s2.

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Most popular questions from this chapter

a. At what height above the earth is the free-fall acceleration 10%of its value at the surface?

b. What is the speed of a satellite orbiting at that height?

A starship is circling a distant planet of radius R. The astronauts find that the free-fall acceleration at their altitude is half the value at the planet’s surface. How far above the surface are they orbiting? Your answer will be a multiple of R.

A 1000 kg satellite and a 2000 kg satellite follow exactly the same orbit around the earth.
a. What is the ratio F1/F2 of the gravitational force on the first satellite to that on the second satellite?
b. What is the ratio a1/a2 of the acceleration of the first satellite to that of the second satellite?

Let’s 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′=2GMr2/r1r1+r2and localid="1651418509687" v2′=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.

b. Consider a localid="1651418519576" 1000kgcommunications satellite that needs to be boosted from an orbit localid="1651418573632" 300kmabove the earth to a geosynchronous orbit localid="1651418578672" 35,900kmabove the earth. Find the velocity localid="1651418584351" v1on the inner circular orbit and the velocity localid="1651418590277" v=1at the low point on the elliptical orbit that spans the two circular orbits.

c. How much work must the rocket motor do to transfer the satellite from the circular orbit to the elliptical orbit?

d. Now find the velocity localid="1651418596735" v=2at the high point of the elliptical orbit and the velocity v2 of the outer circular orbit.

e. How much work must the rocket motor do to transfer the satellite from the elliptical orbit to the outer circular orbit?

f. Compute the total work done and compare your answer to the result of Example localid="1651418602767" 13.6.

Saturn’s moon Titan has a mass of 1.35*1023kgand a radius of2580km. What is the free-fall acceleration on Titan?

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