Gravitation and Satellites: questions with solutions
3 free questions from this chapter (the full set has 30). Marking: Multi-correct: +4 all correct, +1 per correct option if none wrong, −2 if any wrong · Numerical: +4 / 0 · Matching: +3 / −1.
Question 1Level 2 · ExamSingle correct
A thin uniform ring of mass M and radius a lies in the xy-plane with its centre at the origin. Consider points on the z-axis only. The magnitude of the gravitational field on the axis is largest at a distance z0 from the centre, and its value there is E0. The pair (z0,E0) is ______. Exactly one option is correct.
A(a,22a2GM)
B(2a,33a22GM)
C(2a,33a22GM)
D(2a,33a2GM)
Show answer and solution
Answer: B
Every element of the ring is at distance a2+z2 from the axial point; the components perpendicular to the axis cancel, so E(z)=(a2+z2)3/2GMz directed toward the centre.
dzdE=GM(a2+z2)5/2(a2+z2)−3z2=0⇒z0=2a.
Then a2+z02=23a2, so E0=(3a2/2)3/2GM(a/2)=33a22GM.
(At z=a the field is GM/(22a2), which is smaller.) Answer (B).
Difficulty: Hard
Question 2Level 3 · AdvancedSingle correct
A spherically symmetric planet of radius R has density ρ(r)=ρ0(1−Rr) for 0≤r≤R, where r is the distance from its centre and ρ0 is a constant. Let g(r) be the magnitude of the gravitational acceleration at distance r from the centre (0≤r≤R). The value rm at which g(r) is maximum and the ratio g(rm)/g(R) are ______. Exactly one option is correct.
Arm=2R,g(R)g(rm)=34
Brm=32R,g(R)g(rm)=89
Crm=32R,g(R)g(rm)=34
Drm=43R,g(R)g(rm)=1621
Show answer and solution
Answer: C
Mass inside radius r: M(r)=∫0r4πs2ρ0(1−s/R)ds=4πρ0(3r3−4Rr4).
By the shell theorem only this mass matters: g(r)=r2GM(r)=4πGρ0(3r−4Rr2).
g′(r)=0⇒31=2Rr⇒rm=32R.
g(rm)=4πGρ0R(92−91)=94πGρ0R, g(R)=4πGρ0R(31−41)=124πGρ0R.
Ratio =12/9=4/3. Answer (C). (For a uniform planet the maximum would be at the surface; a centrally condensed planet has g rising below the surface.)
Difficulty: Very hard
Question 3Level 3 · AdvancedSingle correct
A planet is a uniform, non-rotating solid sphere of radius R with surface gravitational acceleration g and no atmosphere. A narrow smooth tunnel is drilled along a diameter. A small ball is released from rest at a height R above the surface, directly above one mouth of the tunnel, so that it falls along the line of the tunnel. The time taken by the ball to travel from the mouth of the tunnel (the planet's surface) to the centre of the planet is ______. Exactly one option is correct.
A4πgR
B2πgR
C3πgR
D6πgR
Show answer and solution
Answer: A
Speed at the surface (energy, outside the planet the field is that of a point mass): 21vs2=GM(R1−2R1)⇒vs2=RGM=gR.
Inside a uniform sphere g(r)=gr/R, so the motion along the tunnel is SHM about the centre with ω=g/R. The ball enters at x=R with speed gR, so the amplitude is A=R2+vs2/ω2=2R (larger than R: the ball would overshoot the far mouth if the SHM law held outside).
Write x=Asinωt′, with t′=0 at the centre. At the mouth sinωt′=R/A=1/2, so ωt′=π/4. Time from mouth to centre =4πgR.
Check: speed at the centre =ωA=2gR, which agrees with energy: 21vc2=GM(23R1−2R1)=gR. Answer (A).
Trap: assuming the ball starts the SHM from rest at the surface gives π/2; using amplitude 2R gives π/6.
Difficulty: Very hard
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