Call Now to Set Up Tutoring:

(888) 888-0446

High School Physics : Understanding Universal Gravitation

Study concepts, example questions & explanations for High School Physics

Create An Account

All High School Physics Resources

6 Diagnostic Tests 233 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

← Previous 1 2 3 4 Next →

Example Question #71 : Specific Forces

A 2500kg satellite orbits 3.8*10^8m above the Earth. What is the period of the satellite's orbit?

$m_E=5.97*10^{24}kg$

r_E=6.37*10^5m

$G=6.67*10^{-11}\frac{m^3}{kg*s}$

Possible Answers:

$2.7*10^{-3}s$

5.33*10^7s

5420s

2.84*10^8s

$1.71*10^{11}s$

Correct answer:

$1.71*10^{11}s$

Explanation:

The period describes how long it takes the satellite to make one full orbit. If you go back to the definition of velocity, $v=\frac{\Delta x}{\Delta t}$ , we can apply that to our new circular orbit, in which the distance is equal to the circumference of the circle and the time is equal to the period: $v=\frac{2\pi r}{T}$ . The circumference divided by the period will give us the average velocity.

The problem gives us the radius, but we need to find the tangential velocity. We can do this by first solving for the centripetal acceleration from the centripetal force.

Recognize that the force due to gravity of the Earth on the satellite is the same as the centripetal force acting on the satellite. That means F_c=F_G .

Solve for F_G for the satellite. To do this, use the law of universal gravitation.

$F_G=G\frac{m_1m_2}{r^2}$

Remember that is the distance between the centers of the two objects. That means it will be equal to the radius of the earth PLUS the orbiting distance.

Use the given values for the masses of the objects and distance to solve for the force of gravity.

$F_G=G\frac{m_sm_E}{(r_E+h)^2}$

$F_G=(6.67*10^{-11}\frac{m^3}{kg*s})*\frac{(2500kg)(5.97*10^{24}kg)}{(6.37*10^5m+3.8*10^8m)^2}$

$F_G=(6.67*10^{-11}\frac{m^3}{kg*s^2})*\frac{1.49*10^{28}kg^2}{1.45*10^{17}m^2}$

$F_G=(6.67*10^{-11}\frac{m^3}{kg*s^2})*(1.03*10^{11}\frac{kg^2}{m^2})$

F_G=6.87N

Now that we know the force, we can find the acceleration. Remember that centripetal force is F_c=m*a_c . Set our two forces equal and solve for the centripetal acceleration.

F_G=F_c=m_s*a_c

6.87N=m_s*a_c

6.87N=2500kg*a_c

$2.7*10^{-3}\frac{m}{s^2}=a_c$

Now we can find the tangential velocity, using the equation for centripetal acceleration. Again, remember that the radius is equal to the sum of the radius of the Earth and the height of the satellite!

$a_c=\frac{v^2}{r}$

$2.7*10^{-3}\frac{m}{s^2}=\frac{v^2}{1.45*10^{17}m}$

$(2.7*10^{-3}\frac{m}{s^2})(1.45*10^{17}m)={v^2}$

$2.85*10^{15}\frac{m^2}{s^2}=v^2$

$\sqrt{2.85*10^{15}\frac{m^2}{s^2}}=\sqrt{v^2}$

$5.33*10^7\frac{m}{s}=v$

We now have a value for the tangential velocity, which we can use in the equation for velocity from the beginning to find the period.

$v=\frac{\Delta x}{\Delta t}=\frac{2\pi r}{T}$

$5.33*10^7\frac{m}{s}=\frac{2\pi(1.45*10^{17}m)}{T}$

$T=\frac{2\pi(1.45*10^{17}m)}{5.33*10^7\frac{m}{s}}$

$T=1.71*10^{11}s$

Report an Error

Example Question #31 : Universal Gravitation

Two planets are 3*10^8m apart. If the first planet has a mass of $2.65*10^{20}kg$ and the second has a mass of $6.33*10^{26}kg$ , what is the gravitational force between them?

$G=6.67*10^{-11}\frac{m^3}{kg*s^2}$

Possible Answers:

$2.23*10^{18}N$

$1.24*10^{20}N$

$1.24*10^{19}N$

$4.7*10^{-1}N$

$1.86*10^{30}N$

Correct answer:

$1.24*10^{20}N$

Explanation:

To solve, use Newton's law of universal gravitation:

$F_G=G\frac{m_1m_2}{r^2}$

We are given the values for the mass of each planet, as well as the distance (radius) between them. Using these values and the gravitational constant, we can solve for the force of gravity.

$F_G=G\frac{m_1m_2}{r^2}$

$F_G=(6.67*10^{-11}\frac{m^3}{kg*s^2})*\frac{(2.65*10^{20}kg)*(6.33*10^{26}kg)}{(3*10^8m)^2}$

$F_G=(6.67*10^{-11}\frac{m^3}{kg*s^2})*\frac{1.67745*10^{47}kg^2}{9*10^{16}m^2}$

$F_G=(6.67*10^{-11}\frac{m^3}{kg*s^2})*1.864*10^{30}\frac{kg^2}{m^2}$

$F_G=1.24*10^{20}N$

Report an Error

Example Question #34 : Universal Gravitation

Two planets are 3*10^8m apart. If the first planet has a mass of $2.65*10^{20}kg$ and the second has a mass of $6.33*10^{26}kg$ , what is the acceleration on the smaller planet?

$G=6.67*10^{-11}\frac{m^3}{kg*s^2}$

Possible Answers:

$2\frac{m}{s^2}$

$51048\frac{m}{s^2}$

$0.47\frac{m}{s^2}$

$1.96*10^{-7}\frac{m}{s^2}$

$2.1\frac{m}{s^2}$

Correct answer:

$0.47\frac{m}{s^2}$

Explanation:

Remember that Newton's second law states that F=ma . The force acting upon the planet in question will be the force due to gravity. Once we find that, we can find the acceleration.

To solve for the force, use Newton's law of universal gravitation:

$F_G=G\frac{m_1m_2}{r^2}$

We are given the values for the mass of each planet, as well as the distance (radius) between them. Using these values and the gravitational constant, we can solve for the force of gravity.

$F_G=G\frac{m_1m_2}{r^2}$

$F_G=(6.67*10^{-11}\frac{m^3}{kg*s^2})*\frac{(2.65*10^{20}kg)*(6.33*10^{26}kg)}{(3*10^8m)^2}$

$F_G=(6.67*10^{-11}\frac{m^3}{kg*s^2})*\frac{1.67745*10^{47}kg^2}{9*10^{16}m^2}$

$F_G=(6.67*10^{-11}\frac{m^3}{kg*s^2})*1.864*10^{30}\frac{kg^2}{m^2}$

$F_G=1.24*10^{20}N$

Now that we know the force of gravity, we can use Newton's second law and the mass of the smaller planet to solve for the acceleration of gravity.

F=ma

$1.24*10^{20}N=(2.65*10^{20}kg)*a$

$\frac{1.24*10^{20}N}{2.65*10^{20}kg}=a$

$0.47\frac{m}{s^2}=a$

Report an Error

Example Question #681 : Ap Physics 1

Two planets are 3*10^8m apart. If the first planet has a mass of $2.65*10^{20}kg$ and the second has a mass of $6.33*10^{26}kg$ , what is the acceleration on the larger planet?

$G=6.67*10^{-11}\frac{m^3}{kg*s^2}$

Possible Answers:

$0.47\frac{m}{s^2}$

$1.96*10^{-7}\frac{m}{s^2}$

$5.1*10^3\frac{m}{s^2}$

$5.1*10^6\frac{m}{s^2}$

$2.14\frac{m}{s^2}$

Correct answer:

$1.96*10^{-7}\frac{m}{s^2}$

Explanation:

Remember that Newton's second law states that F=ma . The force acting upon the planet in question will be the force due to gravity. Once we find that, we can find the acceleration.

To solve for the force, use Newton's law of universal gravitation:

$F_G=G\frac{m_1m_2}{r^2}$

We are given the values for the mass of each planet, as well as the distance (radius) between them. Using these values and the gravitational constant, we can solve for the force of gravity.

$F_G=G\frac{m_1m_2}{r^2}$

$F_G=(6.67*10^{-11}\frac{m^3}{kg*s^2})*\frac{(2.65*10^{20}kg)*(6.33*10^{26}kg)}{(3*10^8m)^2}$

$F_G=(6.67*10^{-11}\frac{m^3}{kg*s^2})*\frac{1.67745*10^{47}kg^2}{9*10^{16}m^2}$

$F_G=(6.67*10^{-11}\frac{m^3}{kg*s^2})*1.864*10^{30}\frac{kg^2}{m^2}$

$F_G=1.24*10^{20}N$

Now that we know the force of gravity, we can use Newton's second law and the mass of the larger planet to solve for the acceleration of gravity.

F=ma

$1.24*10^{20}N=(6.33*10^{26}kg)*a$

$\frac{1.24*10^{20}N}{6.33*10^{26}kg}=a$

$1.96*10^{-7}\frac{m}{s^2}=a$

Report an Error

Example Question #682 : Ap Physics 1

An astronaut has a mass of 71kg . He travels to a new planet and observes his weight is 510N on this planet's surface. If the radius of the planet is 6*10^7m , what is the mass of the planet?

$G=6.67*10^{-11}\frac{m^3}{kg*s^2}$

Possible Answers:

$8.34*10^{31}kg$

$2.87*10^{-2}kg$

$6.46*10^{18}kg$

$3.26*10^{22}kg$

$3.87*10^{26}kg$

Correct answer:

$3.87*10^{26}kg$

Explanation:

To solve, use Newton's law of universal gravitation:

$F=G\frac{m_1*m_2}{r^2}$

Remember that the weight of the astronaut is the same as the gravitational force acting between the planet and the astronaut.

$F_g=mg=G\frac{m_1*m_2}{r^2}$

We are given the gravitational constant, the radius of the planet, the mass of the astronaut, and the magnitude of the force generated. Using these values in the universal gravitation equation, we can solve for the mass of the planet.

$510N=(6.67*10^{-11}\frac{m^3}{kg*s^2})*\frac{(71kg)m_2}{(6*10^7m)^2}$

$\frac{510N}{6.67*10^{-11}\frac{m^3}{kg*s^2}}=\frac{(71kg)m_2}{(6*10^7m)^2}$

$7.65*10^{12}\frac{kg^2}{m^2}=\frac{(71kg)m_2}{3.6*10^{15}m^2}$

$(7.65*10^{12}\frac{kg^2}{m^2})(3.6*10^{15}m^2)=(71kg)m_2$

$2.75*10^{28}kg^2=(71kg)m_2$

$3.87*10^{26}=m_2$

Report an Error

Example Question #5 : Universal Gravitation

Two satellites in space, each with equal mass, are 80m apart from each other. If the force of gravity between them is $2.88*10^{-10}N$ , what is the mass of each satellite?

$\small G=6.67*10^{-11}\frac{m^3}{kg\cdot s^2}$

Possible Answers:

2022.44kg

1602.98kg

1.989kg

2405.33kg

166.28kg

Correct answer:

166.28kg

Explanation:

To solve this problem, use Newton's law of universal gravitation:

$F_G=G\frac{m_1m_2}{r^2}$

We are given the value of the force, the distance (radius), and the gravitational constant. We are also told that the masses of the two satellites are equal. Since the masses are equal, we can reduce the numerator of the law of gravitation to a single variable.

$F_G=G\frac{M^2}{r^2}$

Now we can use our give values to solve for the mass.

$2.88*10^{-10}N=(6.67*10^{-11}\frac{m^3}{kg\cdot s^2})(\frac{M^2}{(80m)^2})$

$\frac{2.88*10^{-10}N}{(6.67*10^{-11}\frac{m^3}{kg\cdot s^2})}=\frac{M^2}{6400m^2}$

$4.32\frac{kg^2}{m^2}=\frac{M^2}{6400m^2}$

$6400m^2* 4.32\frac{kg^2}{m^2}=M^2$

27648kg^2=M^2

$\sqrt{27648kg^2}=\sqrt{M^2}$

166.28kg=M

Report an Error

← Previous 1 2 3 4 Next →

View Tutors

Carlos
Certified Tutor

Pikes Peak Community College, Associate in Science, Cyber/Electronic Operations and Warfare.

View Tutors

Rachel
Certified Tutor

University of Arkansas at Little Rock, Bachelor in Arts, Mathematics.

View Tutors

Lama
Certified Tutor

Western University, Bachelor in Arts, Civil Engineering.

All High School Physics Resources

6 Diagnostic Tests 233 Practice Tests Question of the Day Flashcards Learn by Concept

Popular Subjects

ISEE Tutors in Chicago, Chemistry Tutors in Phoenix, LSAT Tutors in Denver, Reading Tutors in Dallas Fort Worth, Reading Tutors in Atlanta, SAT Tutors in Houston, SSAT Tutors in Philadelphia, SSAT Tutors in Chicago, Statistics Tutors in Dallas Fort Worth, GRE Tutors in Atlanta

Popular Courses & Classes

ISEE Courses & Classes in Houston, LSAT Courses & Classes in Houston, GMAT Courses & Classes in Philadelphia, Spanish Courses & Classes in Denver, ISEE Courses & Classes in Miami, SAT Courses & Classes in Miami, ACT Courses & Classes in Seattle, ISEE Courses & Classes in Atlanta, LSAT Courses & Classes in Miami, SAT Courses & Classes in Seattle

Popular Test Prep

GMAT Test Prep in Phoenix, ISEE Test Prep in Denver, SSAT Test Prep in Seattle, LSAT Test Prep in Chicago, SAT Test Prep in Atlanta, ISEE Test Prep in Houston, SSAT Test Prep in Dallas Fort Worth, GMAT Test Prep in Philadelphia, LSAT Test Prep in San Diego, SSAT Test Prep in New York City

DMCA Complaint

If you believe that content available by means of the Website (as defined in our Terms of Service) infringes one or more of your copyrights, please notify us by providing a written notice (“Infringement Notice”) containing the information described below to the designated agent listed below. If Varsity Tutors takes action in response to an Infringement Notice, it will make a good faith attempt to contact the party that made such content available by means of the most recent email address, if any, provided by such party to Varsity Tutors.

Your Infringement Notice may be forwarded to the party that made the content available or to third parties such as ChillingEffects.org.

Please be advised that you will be liable for damages (including costs and attorneys’ fees) if you materially misrepresent that a product or activity is infringing your copyrights. Thus, if you are not sure content located on or linked-to by the Website infringes your copyright, you should consider first contacting an attorney.

Please follow these steps to file a notice:

You must include the following:

A physical or electronic signature of the copyright owner or a person authorized to act on their behalf; An identification of the copyright claimed to have been infringed; A description of the nature and exact location of the content that you claim to infringe your copyright, in \ sufficient detail to permit Varsity Tutors to find and positively identify that content; for example we require a link to the specific question (not just the name of the question) that contains the content and a description of which specific portion of the question – an image, a link, the text, etc – your complaint refers to; Your name, address, telephone number and email address; and A statement by you: (a) that you believe in good faith that the use of the content that you claim to infringe your copyright is not authorized by law, or by the copyright owner or such owner’s agent; (b) that all of the information contained in your Infringement Notice is accurate, and (c) under penalty of perjury, that you are either the copyright owner or a person authorized to act on their behalf.

Send your complaint to our designated agent at:

Charles Cohn Varsity Tutors LLC
101 S. Hanley Rd, Suite 300
St. Louis, MO 63105

Or fill out the form below:

Do not fill in this field

Contact Information

* Full legal name

Company name

* Phone number

* Email address

Address

* Address1

Address2

* City

* State

* Zipcode

* Country

Complaint Details

* URL of copyrighted work (where your original material is located)

* Please describe the copyrighted work so that it may be easily identified

I am the owner, or an agent authorized to act on behalf of the owner of an exclusive right that is allegedly infringed.

I have a good faith belief that the use of the material in the manner complained of is not authorized by the copyright owner, its agent, or the law.

This notification is accurate.

I acknowledge that there may be adverse legal consequences for making false or bad faith allegations of copyright infringement by using this process.

* Signature (your digital signature is legally binding)

HIGH SCHOOL

GRADUATE SCHOOL

K-8

Search 50+ Tests

math tutoring

science tutoring

foreign languages

elementary tutoring

other

Search 350+ Subjects

High School Physics : Understanding Universal Gravitation

Study concepts, example questions & explanations for High School Physics

All High School Physics Resources

Example Questions

Example Question #71 : Specific Forces

Example Question #31 : Universal Gravitation

Example Question #34 : Universal Gravitation

Example Question #681 : Ap Physics 1

Example Question #682 : Ap Physics 1

Example Question #5 : Universal Gravitation

All High School Physics Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: