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Feedback Comments on Gravitation

by Ron Kurtus

A total of 63 comments and questions have been sent in on Gravitation. They are listed according to date.

You can read them to further your understanding of the subject.



List of most recent 10 letters

Topic

Title

Country

Force Between Two Objects Breaking mass in parts Tanzania
 
Cavendish Experiment Changing parameters in Cavendish experiment Egypt
 
Force Between Two Objects Where are the attractive forces equal? USA
 
General Viewpoint on gravitation and Big Bang theory Bangladesh
 
Force Between Two Objects What do the signs mean Australia
 
Force Between Two Objects Gravitational force USA
 
Force Between Two Objects Correction on exponent UK
 
Gravitational Escape Velocity Derivation Escape velocity equation seems wrong Nigeria
 
Force Between Two Objects Gravitation miscalculation USA
 
Tides Tidal acceleration equation USA
 

Next 10

 





Breaking mass in parts

Topic: Force Between Two Objects

Question

October 18, 2017

a mass M is broken into two parts "m" and (M-m) related so that of gravitational between the two parts is maximum??

erickson - Tanzania

28292

Answer

The force is proportional to m(M-m) or (mM - m^2).

If you try values of m = 0.1M, m = 0.2M, m = 0.3M, m = 0.5M, m = 0.6M and so on, you will find that the maximum value of m(M-m) is at m = 0.5M. In other words, the maximum gravitational force is when both pieces are the same size.

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Changing parameters in Cavendish experiment

Topic: Cavendish Experiment

Question

October 11, 2017

If we change
1 - Diameters of the balls
2 - masses of the balls
3 - metals of the balls
4- L torsion bar length
if you change the previous parameters will G still constant factor???!!!
I wish your answer and Thank you very much.

Anwer - Egypt

28281

Answer

You can change the parameters and perform the Cavendish Experiment to get the value of the gravitation constant G. However, the values need to be optimized to get the most accurate result. For example, if the masses of the balls were decreased, the resulting value for G would not be as accurate and there might be some errors in the measurement.

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Where are the attractive forces equal?

Topic: Force Between Two Objects

Question

August 5, 2017

At what distance from the earth should an object lie between the earth and the sun such that the attractive force of the earth on it is equal the attractive force the sun on it , given that the distance between the earth and sun is 150million Km and the mass of the sun =3.24mass of earthh

- USA

28163

Answer

The point where the attractive forces are equal is at the center of mass (CM) between the Earth and the Sun. See CM between two spheres for the equation.

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Viewpoint on gravitation and Big Bang theory

Topic: General

Question

June 19, 2017

Near about 14 billion years ago Allah/God the big bang and started evolution, now 7 divisible gravitation worlds.

“Gravitational worlds, they are moving or changing the orbit with their all family members depending on the nuclear of each other”. In this way- There has been incidence of full view of the universe at the very exact copy/invitation of the visible diagram solar system in the gravitation world. The universe has seven divisible, the first one of which is unit and the rest six are collective. The collective six gravitation are three invisible and the rest three are visible. The visible gravitation worlds are stars, planets and minor planets. The invisible gravitation worlds are black holes of various energy levels, the first one of which is the primordial black hole or Allah/God of the nuclear fusion of the universe. There have been collective six gravitation divisible of the big bang in future from the big black hole or Allah/God. At the first phase or level of the divisible is the black hole of the nuclear fusion of the clusters of galaxies or black hole named quasar. The second phase gravitation world of the quasars in future are of the black hole named pulsar or nuclear fusion of the cluster of galaxies. The third gravitation world of the pulsars in future is of the last black holes or nucleus of galaxy or the nuclear fusion of the constellations.

New Discovery of the Universe: https://shahidurrahmansikder.wordpress.com/tag/god/

Pope: The most important news- about the God, Big Bang & Evolution at https://plus.google.com/104669722445739033329/posts/MQ9wQejCRLQ

Shahidur Rahman - Bangladesh

28099

Answer

Thanks for the information and interesting viewpoint on gravitation.

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What do the signs mean

Topic: Force Between Two Objects

Question

February 19, 2017

Hey I'm Aric and I'm trying to work out the gravitational attraction between the Sun and the Earth for homework
I just wanna know what the times, divide, plus and minus mean
Like as in;
* = Times?
/ = Divide/Fraction
- = Minus
and so on
Thanks

Aric - Australia

27888

Answer

The equation F = GMm/R^2 means G times M times m divided by R squared or R times R.

To answer your question:
a*b means a times b or multiply a times b
a/b is a divided by b
a - b is a minus b or subtract b from a

The big thing is to find the mass of the Sun and of the Earth, as well as the distance between them.

I hope that helps. Best wishes in getting the problem solved.

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Gravitational force

Topic: Force Between Two Objects

Question

January 28, 2017

1. Two objects attract each other with a force of 1×10^-8 N when separated by 20 cm. If the total mass of the objects is 5 kg what is the mass of each?

2. A spacecraft is on a journey to the moon. At what point between these two bodies does the gravitational force exerted on the spacecraft by the earth exactly balance gravitational force divided by the moon? At this point what is the net gravitational force on the spacecraft? assuming no other gravitational forces act.
I'd really love some help figuring these out!! Thank you!

Emily - USA

27831

Answer

In the gravitational force equation, let "m" equal the mass of one object and "5 - m" be the mass of the other object. Thus, the equation is:

F = G(m)(m-5)/R^2
Thus: F(R^2)/G = m^2 - 5m
Use the quadratic formula to solve for m.
Note that R must be in meters.

See Law of Universal Gravitation for more information.

The second problem is solved by using the forces from the Earth and from the Moon on the spacecraft: GmM/r^2 = GmE/R^2
where m is the mass of the spacecraft, M is the mass of the Moon, r is the distance from the Moon to the spacecraft, E is the mass of the Earth, R is the distance from the Earth to the spacecraft. Also D = r + R the distance between the Earth and the Moon. That involves a lot of Algebra.

The net force is zero, since the gravitational forces cancel out.

I hope that helps.

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Correction on exponent

Topic: Force Between Two Objects

Question

February 25, 2016

Hi, Great site, but your page with the calc of the gravitational force between two people is wrong at the end when you convert to standard scientific notation. It should be 0.9x10-6, not -16

- UK

26809

Answer

Thanks for pointing that out to me. I made a goof on handling the exponents. I corrected the Gravitational Force Between Two Objects page.

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Escape velocity equation seems wrong

Topic: Gravitational Escape Velocity Derivation

Question

January 18, 2016

you wrote that ve = ??(2GM/Ri) but in my textbook I saw that ve = ?2gR

valentine - Nigeria

26708

Answer

The equation ve = SQRT(2GM/Ri) is the escape velocity from the gravitation of any very large object, such as the Sun.

However, you can simplify the equation for the escape from the Earth as v = SQRT(2gR), where g is the acceleration due to gravity and R is the radius of the Earth. See Escape Velocity from Gravity - Common Escape Velocities paragraph.

I hope that clarifies things.

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Gravitation miscalculation

Topic: Force Between Two Objects

Question

January 4, 2016

The boy is 165lbs has a weight of about 740 Newtons and a mass of 70-kg not 7-kg.

Steve - USA

26661

Answer

Thanks for noticing the miscalculation..

There is so much confusing between kg-mass and kg-force, that I put everything in terms of mass in the Gravitational Force Between Two Objects calculations.

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Tidal acceleration equation

Topic: Tides

Question

November 5, 2015

Ron,

I bought a copy of your Gravity and Gravitation book. Please answer this question:

In NASA Science News, Frank Reed said the following:

I calculated the tidal influence of each planet at the Earth over the course of several centuries using standard almanac algorithms (only a few decades are presented here). The tidal acceleration is proportional to the mass of the planet, inversely proportional to the distance cubed, and also depends on direction.

Ron, my question is, "Why is the tidal acceleration inversely proportional to the distance cubed, instead of to the distance squared?"

Thank you very much!

Dick - USA

26500

Answer

The whole subject of tides can get quite complex. The tidal acceleration is defined as the gradient or change in force per change in distance.

If you take the derivative of the gravitational force with respect to distance, you get: dF/dR = d(GMm/R^2) = 2GMm/R^3.

I hope that clears things up.

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