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Chapter 13: Q. 9 (page 353)

What is the free-fall acceleration at the surface of (a) the moon and (b) Jupiter?

Short Answer

Expert verified

The free-fall acceleration at the surface of (a) the moon is 1.62m/s2; and (b) Jupiter is24.8m/s2.

Step by step solution

01

Given information (a).

Universal gravitational constant: G=6.67×10-11Nm2/kg2.

Mass of moon: Mm=7.34×1024kg.

Radius of moon rm=1.74×106m.

02

Calculation (a).

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

The free-fall acceleration at the on the surface of moon is :

localid="1648481120225" gm=GMmrm2=(6.67×10-11Nm2/kg2)(7.34×1024kg)(1.74×106m)2.=1.62m/s2.

03

Final answer (a).

The free-fall acceleration at the on the surface of moon is1.62m/s2.

04

Given information (b).

Universal gravitational constant: G=6.67×10-11Nm2/kg2.

Mass of Jupiter: Mj=1.90×1027kg.

Radius of Jupiter: rj=6.99×107m.

05

Calculation (b).

The free-fall acceleration at the surface of Jupiter is :

gj=GMjrj2=(6.67×10-11Nm2/kg2)(1.9×1027kg)(6.99×107m)2=24.8m/s2.

06

Final answer (b).

The free-fall acceleration at the surface of Jupiter is24.8m/s2.

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