Ch 08: Dynamics II: Motion in a Plane

Knight Calc - Physics for Scientists and Engineers 5th Edition

Knight Calc5th EditionPhysics for Scientists and EngineersISBN: 9780137344796Not the one you use?Change textbook

Knight Calc 5th Edition

Ch 08: Dynamics II: Motion in a Plane

Problem 14

Chapter 8, Problem 14

A satellite orbiting the moon very near the surface has a period of 110 min. What is free-fall acceleration on the surface of the moon? Astronomical data are inside the back cover of the book.

Verified step by step guidance

Step 1: Start by understanding the relationship between the orbital period and the gravitational force. The centripetal force required to keep the satellite in orbit is provided by the gravitational force. Use the formula for centripetal force:

F = \frac{m^{2}}{r}

, where

m

is the mass of the satellite,

v

is its orbital velocity, and

r

is the radius of the orbit.

Step 2: Relate the orbital velocity to the period of the satellite. The velocity

v

can be expressed as

v = \frac{2 π r}{T}

, where

T

is the orbital period. Substitute this expression for

v

into the centripetal force equation.

Step 3: Use Newton's law of gravitation to express the gravitational force:

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

, where

G

is the gravitational constant,

M

is the mass of the moon, and

r

is the radius of the moon. Set the gravitational force equal to the centripetal force.

Step 4: Simplify the equation to solve for the free-fall acceleration

g

on the surface of the moon. The free-fall acceleration is given by

g = \frac{G M}{r^{2}}

. Use the relationship between the orbital period and the radius to eliminate

M

and express

g

in terms of known quantities.

Step 5: Substitute the given values for the orbital period

T

(110 minutes converted to seconds), the radius of the moon (from the astronomical data), and the gravitational constant

G

. Perform the algebraic manipulations to find the free-fall acceleration

g

Verified video answer for a similar problem:

This video solution was recommended by our tutors as helpful for the problem above.

Video duration:

Was this helpful?

Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Orbital Period

The orbital period is the time it takes for an object to complete one full orbit around a celestial body. In this case, the satellite's period of 110 minutes indicates how long it takes to circle the moon. This period is crucial for calculating the gravitational force acting on the satellite, which is directly related to the free-fall acceleration experienced on the moon's surface.

Gravitational Acceleration

Gravitational acceleration is the acceleration experienced by an object due to the gravitational force exerted by a massive body, such as the moon. On the moon, this value is significantly lower than on Earth, approximately 1.62 m/s². Understanding this concept is essential for determining how the satellite's orbital characteristics relate to the moon's gravitational pull.

Newton's Law of Universal Gravitation

Newton's Law of Universal Gravitation states that every mass attracts every other mass with a force that is proportional to the product of their masses and inversely proportional to the square of the distance between their centers. This law is fundamental for calculating the gravitational force acting on the satellite and, consequently, the free-fall acceleration on the moon's surface, as it provides the relationship needed to derive these values.

Recommended video:

Guided course

04:05

Universal Law of Gravitation

Related Practice

Textbook Question

A car drives over the top of a hill that has a radius of 50 m. What maximum speed can the car have at the top without flying off the road?

views

Textbook Question

Three satellites orbit a planet of radius R, as shown in FIGUREEX13.24. Satellites S₁ and S₃ have mass m. Satellite S₂ has mass 2m. Satellite S₁ orbits in 250 minutes and the force on S₁ is 10,000 N.(b) What are the forces of S₂ and S₃?

views

Textbook Question

Communications satellites are placed in circular orbits where they stay directly over a fixed point on the equator as the Earth rotates. These are called geosynchronous orbits. The altitude of a geosynchronous orbit is 3.58 x 10⁷ m (approximately 22,00 miles). Astronomical data are inside the back cover of the book. What is the weight of a 2000 kg satellite in a geosynchronous orbit?

Textbook Question

It is proposed that future space stations create an artificial gravity by rotating. Suppose a space station is constructed as a 1000-m-diameter cylinder that rotates about its axis. The inside surface is the deck of the space station. What rotation period will provide 'normal' gravity?

A 5.0 g coin is placed 15 cm from the center of a turntable. The coin has static and kinetic coefficients of friction with the turntable surface of μ_s = 0.80 and μ_k = 0.50. The turntable very slowly speeds up to 60 rpm. Does the coin slide off?

views