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

Example Question #41 : Forces

Two satellites in space, each with a mass of 2000kg , are 1500m apart from each other. What is the force of gravity between them?

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

Possible Answers:

$1.19*10^{-10}N$

$2.66*10^{-11}N$

$8.87*10^{-10}N$

$6.22*10^{-8}N$

$3.12*10^{-1}N$

Correct answer:

$1.19*10^{-10}N$

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 constant, as well as the satellite masses and distance (radius). Using these values we can solve for the force.

$F_G=(6.67*10^{-11}\frac{m^3}{kg\cdot s^2})(\frac{(2000kg) (2000kg)}{(1500m)^2})$

$F_G=(6.67*10^{-11}\frac{m^3}{kg\cdot s^2})(\frac{4000000kg^2}{2250000m^2})$

$F_G=1.19*10^{-10}N$

Report an Error

Example Question #3 : Universal Gravitation

Two satellites in space, each with a mass of 1723kg , are 890m apart from each other. What is the force of gravity between them?

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

Possible Answers:

$1.1*10^{3}N$

$1.25*10^{-19}N$

$2.5*10^{-20}N$

$2.2*10^{8}N$

$2.5*10^{-10}N$

Correct answer:

$2.5*10^{-10}N$

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 constant, as well as the satellite masses and distance (radius). Using these values we can solve for the force.

$F_G=(6.67*10^{-11}\frac{m^3}{kg\cdot s^2})(\frac{(1723kg )(1723kg)}{(890m)^2})$

$F_G=(6.67*10^{-11}\frac{m^3}{kg\cdot s^2})(\frac{2968729kg^2}{792100m^2})$

$F_G=2.5*10^{-10}N$

Report an Error

Example Question #4 : Universal Gravitation

Two asteroids in space are in close proximity to each other. Each has a mass of $6.69*10^{15}kg$ . If they are 100,000m apart, what is the gravitational force between them?

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

Possible Answers:

$3.01*10^{-8}N$

$2.99*10^{11}N$

$5.98*10^{12}N$

$1.11*10^{30}N$

$1.49*10^{-32}N$

Correct answer:

$2.99*10^{11}N$

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 constant, as well as the asteroid masses and distance (radius). Using these values we can solve for the force.

$F_G=(6.67*10^{-11}\frac{m^3}{kg\cdot s^2})(\frac{(6.69*10^{15}kg )(6.69*10^{15}kg )}{(100,000m)^2})$

$F_G=6.67*10^{-11}\frac{m^3}{kg\cdot s^2}(\frac{4.48*10^{31}kg^2}{10*10^9m^2})$

$F_G=6.67*10^{-11}\frac{m^3}{kg\cdot s^2}(4.48*10^{21}\frac{kg^2}{m^2})$

$F_G=2.99*10^{11}N$

Report an Error

Example Question #1 : Universal Gravitation

Two asteroids in space are in close proximity to each other. Each has a mass of $6.69*10^{15}kg$ . If they are 100,000m apart, what is the gravitational acceleration that they experience?

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

Possible Answers:

$6.69*10^{15}\frac{m}{s^2}$

$3.89*10^{10}\frac{m}{s^2}$

$4.47*10^{-5}\frac{m}{s^2}$

$5.12*10^{4}\frac{m}{s^2}$

$2.99*10^{11}\frac{m}{s^2}$

Correct answer:

$4.47*10^{-5}\frac{m}{s^2}$

Explanation:

Given that F=ma , we already know the mass, but we need to find the force in order to solve for the acceleration.

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

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

We are given the constant, as well as the satellite masses and distance (radius). Using these values we can solve for the force.

$F_G=6.67*10^{-11}\frac{m^3}{kg\cdot s^2}(\frac{(6.69*10^{15}kg)(6.69*10^{15}kg)}{(100,000m)^2})$

$F_G=6.67*10^{-11}\frac{m^3}{kg\cdot s^2} (\frac{4.48*10^{31}kg^2}{10*10^9m^2})$

$F_G=6.67*10^{-11}\frac{m^3}{kg\cdot s^2}(4.48*10^{21}\frac{kg^2}{m^2})$

$F_G=2.99*10^{11}N$

Now we have values for both the mass and the force, allowing us to solve for the acceleration.

F=ma

$2.99*10^{11}N=(6.69*10^{15}kg)a$

$\frac{2.99*10^{11}N}{6.69*10^{15}kg}{}=a$

$4.47*10^{-5}\frac{m}{s^2}=a$

Report an Error

Example Question #1 : Universal Gravitation

Two asteroids, one with a mass of $7.12*10^{18}kg$ and the other with mass 5.33*10^8kg , are $10*10^{10}m$ apart. What is the gravitational force on the LARGER asteroid?

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

Possible Answers:

$2.53*10^{-5}N$

$3.55*10^{-6}N$

3.79*10^5N

$4.61*10^{-10}N$

$4.74*10^{-6}N$

Correct answer:

$2.53*10^{-5}N$

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 constant, as well as the asteroid masses and distance (radius). Using these values we can solve for the force.

$F_G=6.67*10^{-11}\frac{m^3}{kg\cdot s^2}( \frac{(7.12*10^{18}kg)(5.33*10^{8}kg)}{(10*10^{10}m)^2})$

$F_G=6.67*10^{-11}\frac{m^3}{kg\cdot s^2} (\frac{3.79*10^{27}kg}{10*10^{21}m^2})$

$F_G=6.67*10^{-11}\frac{m^3}{kg\cdot s^2} (3.79*10^5\frac{kg^2}{m^2})$

$F_G=2.53*10^{-5}N$

It actually doesn't matter which asteroid we're looking at; the gravitational force will be the same. This makes sense because Newton's 3rd law states that the force one asteroid exerts on the other is equal in magnitude, but opposite in direction, to the force the other asteroid exerts on it.

Report an Error

Example Question #3 : Universal Gravitation

Two asteroids, one with a mass of $7.12*10^{18}kg$ and the other with mass 5.33*10^8kg are $10*10^{10}m$ apart. What is the gravitational force on the SMALLER asteroid?

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

Possible Answers:

$4.74*10^{-14}N$

3.79*10^5N

$3.55*10^{-24}N$

$2.53*10^{-5}N$

$1.11*10^{12}N$

Correct answer:

$2.53*10^{-5}N$

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 constant, as well as the asteroid masses and distance (radius). Using these values we can solve for the force.

$F_G=6.67*10^{-11}\frac{m^3}{kg\cdot s^2}( \frac{(7.12*10^{18}kg)(5.33*10^{8}kg)}{(10*10^{10}m)^2})$

$F_G=6.67*10^{-11}\frac{m^3}{kg\cdot s^2} (\frac{3.79*10^{27}kg}{10*10^{21}m^2})$

$F_G=6.67*10^{-11}\frac{m^3}{kg\cdot s^2} (3.79*10^5\frac{kg^2}{m^2})$

$F_G=2.53*10^{-5}N$

Report an Error

Example Question #41 : Forces

Two asteroids, one with a mass of $7.12*10^{18}kg$ and the other with mass 5.33*10^8kg are $10*10^{10}m$ apart. What is the acceleration of the SMALLER asteroid?

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

Possible Answers:

$3.79*10^5\frac{m}{s^2}$

$9.41*10^{-7}\frac{m}{s^2}$

$2.53*10^{-5}\frac{m}{s^2}$

$6.11*10^{-24}\frac{m}{s^2}$

$4.74*10^{-14}\frac{m}{s^2}$

Correct answer:

$4.74*10^{-14}\frac{m}{s^2}$

Explanation:

Given that Newton's second law is F=ma , we can find the acceleration by first determining the force.

To find the gravitational force, use Newton's law of universal gravitation:

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

We are given the constant, as well as the asteroid masses and distance (radius). Using these values we can solve for the force.

$F_G=6.67*10^{-11}\frac{m^3}{kg\cdot s^2}( \frac{(7.12*10^{18}kg)(5.33*10^{8}kg)}{(10*10^{10}m)^2})$

$F_G=6.67*10^{-11}\frac{m^3}{kg\cdot s^2} (\frac{3.79*10^{27}kg}{10*10^{21}m^2})$

$F_G=6.67*10^{-11}\frac{m^3}{kg\cdot s^2} (3.79*10^5\frac{kg^2}{m^2})$

$F_G=2.53*10^{-5}N$

We now have values for both the mass and the force. Using the original equation, we can now solve for the acceleration.

F=ma

$2.53*10^{-5}N=(5.33*10^8kg)a$

$\frac{2.53*10^{-5}N}{5.33*10^8kg}{}=a$

$4.74*10^{-14}\frac{m}{s^2}=a$

Report an Error

Example Question #42 : Forces

Two asteroids, one with a mass of $7.12*10^{18}kg$ and the other with mass 5.33*10^8kg are $10*10^{10}m$ apart. What is the acceleration of the LARGER asteroid?

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

Possible Answers:

$7.12*10^{-18}\frac{m}{s}$

$4.99*10^{-7}\frac{m}{s^2}$

$1.63*10^{-12}\frac{m}{s^2}$

$3.55*10^{-24}\frac{m}{s}$

$3.92*10^{-42}\frac{m}{s^2}$

Correct answer:

$3.55*10^{-24}\frac{m}{s}$

Explanation:

Given that Newton's second law is F=ma , we can find the acceleration by first determining the force.

To find the gravitational force, use Newton's law of universal gravitation:

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

We are given the constant, as well as the asteroid masses and distance (radius). Using these values we can solve for the force.

$F_G=6.67*10^{-11}\frac{m^3}{kg\cdot s^2}( \frac{(7.12*10^{18}kg)(5.33*10^{8}kg)}{(10*10^{10}m)^2})$

$F_G=6.67*10^{-11}\frac{m^3}{kg\cdot s^2} (\frac{3.79*10^{27}kg}{10*10^{21}m^2})$

$F_G=6.67*10^{-11}\frac{m^3}{kg\cdot s^2} (3.79*10^5\frac{kg^2}{m^2})$

$F_G=2.53*10^{-5}N$

We now have values for both the mass and the force. Using the original equation, we can now solve for the acceleration.

F=ma

$2.53*10^{-5}N=(7.12*10^{18}kg)a$

$\frac{2.53*10^{-5}N}{7.12*10^{18}kg}{}=a$

$3.55*10^{-24}\frac{m}{s^2}=a$

Report an Error

Example Question #41 : Forces

Two satellites are a distance from each other in space. If one of the satellites has a mass of and the other has a mass of , which one will have the smaller acceleration?

Possible Answers:

Neither will have an acceleration

They will both have the same acceleration

We need to know the value of the masses to solve

Correct answer:

Explanation:

The formula for force and acceleration is Newton's 2nd law: F=ma . We know the mass, but first we need to find the force:

For this equation, use the law of universal gravitation:

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

We know from the first equation that a force is a mass times an acceleration. That means we can rearrange the equation for universal gravitation to look a bit more like that first equation:

$F=G\frac{m_1m_2}{r^2}$ can turn into: $F=m_2*\frac{G*m_1}{r^2}$ and $F=m_1*\frac{G*m_2}{r^2}$ , respectively.

We know that the forces will be equal, so set these two equations equal to each other:

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

The problem tells us that m_2=2m_1

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

Let's say that $\frac{G*m_1}{r^2}=a$ to simplify.

m_1*2a=2m_1*a

As you can see, the acceleration for m_1 is twice the acceleration for m_2 . Therefore the mass will have the smaller acceleration.

Report an Error

Example Question #51 : Forces

An asteroid with a mass of $10*10^{15}kg$ approaches the Earth. If they are 250,000,000m apart, what is the gravitational force exerted by the asteroid on the Earth?

$\small M_{Earth}=5.972*10^{24}kg$

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

Possible Answers:

$1.59*10^{22}N$

$6.7*10^{30}N$

$6.37*10^{13}N$

$1.27*10^{14}N$

Correct answer:

$6.37*10^{13}N$

Explanation:

For this question, use the law of universal gravitation:

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

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

$F=(6.67*10^{-11}\frac{m^3}{kg*s^2})*\frac{(10*10^{15}kg)*(5.972*10^{24}kg)}{(250,000,000m)^2}$

$F=(6.67*10^{-11}\frac{m^3}{kg*s^2})*\frac{5.972*10^{40}kg^2}{6.25*10^{16}m^2}$

$F=(6.67*10^{-11}\frac{m^3}{kg*s^2})*(9.5552*10^{23}\frac{kg^2}{m^2})$

$F=6.37*10^{13}N$

This force will apply to both objects in question. As it turns out, it does not matter which mass we're looking at; the force of gravity on each mass will be the same. This is supported by Newton's third law.

Report an Error

View Tutors

Mayowa
Certified Tutor

Logan University, Bachelor of Science, Biology, General.

View Tutors

Joy
Certified Tutor

Boston University, Bachelor in Arts, Conservation Biology.

View Tutors

Rebekah
Certified Tutor

Swarthmore College, Bachelor, Psychology & Linguistics. Pace University, Master's/Graduate, School Psychology.

All High School Physics Resources

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

Popular Subjects

MCAT Tutors in San Francisco-Bay Area, Algebra Tutors in Boston, LSAT Tutors in New York City, MCAT Tutors in San Diego, GRE Tutors in Washington DC, ACT Tutors in Miami, Physics Tutors in Los Angeles, ISEE Tutors in Chicago, Computer Science Tutors in Dallas Fort Worth, Physics Tutors in Washington DC

Popular Courses & Classes

SSAT Courses & Classes in Houston, GRE Courses & Classes in San Diego, SAT Courses & Classes in Dallas Fort Worth, MCAT Courses & Classes in Boston, GMAT Courses & Classes in Houston, LSAT Courses & Classes in Dallas Fort Worth, GMAT Courses & Classes in Phoenix, GMAT Courses & Classes in Seattle, ACT Courses & Classes in Denver, SAT Courses & Classes in New York City

Popular Test Prep

GMAT Test Prep in Phoenix, SSAT Test Prep in Chicago, MCAT Test Prep in Boston, LSAT Test Prep in Philadelphia, LSAT Test Prep in Boston, LSAT Test Prep in Denver, ACT Test Prep in Washington DC, GMAT Test Prep in Seattle, ACT Test Prep in Denver, GRE Test Prep in Boston

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 #41 : Forces

Example Question #3 : Universal Gravitation

Example Question #4 : Universal Gravitation

Example Question #1 : Universal Gravitation

Example Question #1 : Universal Gravitation

Example Question #3 : Universal Gravitation

Example Question #41 : Forces

Example Question #42 : Forces

Example Question #41 : Forces

Example Question #51 : Forces

All High School Physics Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: