Chapter 8: Q.16 (page 199)

A $9.4 \times 10^{21} k g$ moon orbits a distant planet in a circular orbit of radius $1.5 \times 10^{8} m$ . It experiences a $1.1 \times 10^{19} N$ gravitational pull from the planet. What is the moon’s orbital period in earth days?

Short Answer

Expert verified

The moon's orbital period in earth days is $= 26$ days

Step by step solution

Given information

Moon orbit's mass $= 9.4 \times 10^{21} k g$

Moon orbit's radius $= 1.5 \times 10^{8} m$

Gravitational pull $= 1.1 \times 10^{19} N$

Explanation

Gravitational pull $= F = \frac{G M m}{r^{2}}$

localid="1649397302533" $F = 1.1 \times 10^{19} N$

localid="1649397341555" $G = 6.67 \times 10^{- 11} N m^{2} k g^{- 2}$

$m = 9.4 \times 10^{21} kg$

localid="1649397362538" $r = 1.5 \times 10^{8} m$

Substitute the values

localid="1649397565238" $1.1 \times 10^{19} N = (\frac{(6.67 \times 10^{- 11} N m^{2} k g^{- 2}) (9.4 \times 10^{21} k g)}{{(1.5 \times 10^{8} m)}^{2}})$

$\Rightarrow M =$ Mass of the distant planet $= 3.947 \times 10^{23} kg$

Moon's orbital period in earth days is given by the expression,

$T = 2 π \sqrt{\frac{r^{3}}{G (M + m)}}$

localid="1649397468079" $= 2 \times 3.14 \sqrt{(\frac{{(1.5 \times 10^{8} m)}^{3}}{(6.67 \times 10^{- 11} N m^{2} k g^{- 2}) [(3.947 \times 10^{23} k g) + (9.4 \times 10^{21} k g)]})}$

$= 2248240 s$

localid="1649397499772" $= (\frac{2248240 s}{3600 s \times 24 s}) days$

$= 26$ days

Final answer

The moon’s orbital period in earth days is

$= 26$ days

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

A $9.4 \times 10^{21} k g$ moon orbits a distant planet in a circular orbit of radius $1.5 \times 10^{8} m$ . It experiences a $1.1 \times 10^{19} N$ gravitational pull from the planet. What is the moon’s orbital period in earth days?

Short Answer

Step by step solution

Given information

Explanation

Final answer

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Physics Textbooks

Medical Physics

Geometrical and Physical Optics

Wave Optics

Famous Physicists

Fields in Physics

Engineering Physics

Study anywhere. Anytime. Across all devices.

A 9.4×1021kgmoon orbits a distant planet in a circular orbit of radius 1.5×108m. It experiences a 1.1×1019Ngravitational pull from the planet. What is the moon’s orbital period in earth days?

Short Answer

Step by step solution

Given information

Explanation

Final answer

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Physics Textbooks

Medical Physics

Geometrical and Physical Optics

Wave Optics

Famous Physicists

Fields in Physics

Engineering Physics

Study anywhere. Anytime. Across all devices.

A $9.4 \times 10^{21} k g$ moon orbits a distant planet in a circular orbit of radius $1.5 \times 10^{8} m$ . It experiences a $1.1 \times 10^{19} N$ gravitational pull from the planet. What is the moon’s orbital period in earth days?