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# Feedback Comments on Gravitation

by Ron Kurtus

A total of **65** 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 next 10 letters

## Topic |
## Title |
## Country |

Force Between Two Objects |
Gravitation miscalculation | USA |

Tides |
Tidal acceleration equation | USA |

Force Between Two Objects |
Graviational pull on a girl | USA |

Force Between Two Objects |
Should the force be inifinte is the separation is zero? | USA |

Gravitational Escape Velocity for a Black Hole |
What is the equation for a Black Hole? | USA |

Force Between Two Objects |
Force of the earth on a girl | Australia |

Cavendish Experiment |
Difference between gravity and gravitation | Kenya |

Cavendish Experiment |
Find the value of G | India |

Force Between Two Objects |
Which will hit the earth first? | USA |

Gravitation |
Calculating weight of Earth twin | USA |

## 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.

## 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.

## Graviational pull on a girl

Topic: **Force Between Two Objects**

### Question

October 27, 2015

Um when You say like the gravitational pull with a girl and moon don't you mean a mermaid when u say that?

Summer - **USA**

26483

### Answer

The idea there is that the gravitational attraction from the Moon pulls on all objects and people on the Earth. That pull is also the reason for the tides.

## Should the force be inifinte is the separation is zero?

Topic: **Force Between Two Objects**

### Question

September 28, 2015

How is it that if the distance between the 2 objects is zero that you can pull the objects apart. Hypothetically speaking the force between the 2 objects should be infinite if the distance between them is zero.

Karim - **USA**

26412

### Answer

The distance between objects is measured from their centers. Thus, since they have some size, and if they are touching, there still would be a separation.

If the particles were points, and they were touching, there separation would be zero and the force would be infinite.

One thing to note is that when particles approach an extremely small size, Quantum Wave Mechanics takes hold, and the particles become considered as waveforms. Different rules then apply.

## What is the equation for a Black Hole?

Topic: **Gravitational Escape Velocity for a Black Hole**

### Question

July 15, 2015

i know light can't escape a black hole but can you give me an example of what the speed would have to be to escape a black hole say if our sun was compressed to the size necessary to create a black hole. I am not a physicist. I saw your equations but I don't know how to calculate the theoretical speed to escape a black hole. Thanks.

William - **USA**

26174

### Answer

The gravitational force of a Black Hole is so great that not even light or some particle traveling at or near the speed of light can escape. According to the theory of Relativity, nothing can exceed the speed of light.

The escape velocity equation is s = ?(2GM/R). If the value of s > c (s is greater than the speed of light) the object is a Black Hole and nothing can escape it.

## Force of the earth on a girl

Topic: **Force Between Two Objects**

### Question

May 23, 2015

The force of the earth on the girl is mg = 9.8 x 50 = 490N

David - **Australia**

26026

### Answer

Although you can calculate the gravitational force between a person and the Earth, it is easier to use the equation F = mg, which actually gives the weight of the person on Earth.

See: Mass, Weight and Gravity for more information.

## Difference between gravity and gravitation

Topic: **Cavendish Experiment**

### Question

February 27, 2015

What is the difference between force of gravity and universal gravitational constant?

CLEOPHAS - **Kenya**

25796

### Answer

Gravity is a force pulling objects toward the Earth. It is an approximation of gravitation, which is the attraction between objects.

In the Cavendish Experiment, the gravitational force between two masses is measured to find the constant G.

See Gravity and Gravitation for more information on those subjects.

## Find the value of G

Topic: **Cavendish Experiment**

### Question

February 7, 2015

How we find the value of G by ths equation G=kdr^2/4MmlD

Asif - **India**

25753

### Answer

The lesson Cavendish Experiment to Measure Gravitational Constant explains how to get to the value of G, which is stated toward the end of the lesson.

## Which will hit the earth first?

Topic: **Force Between Two Objects**

### Question

January 28, 2015

which will hit the earth first: a bullet fired from 6 feet high parallel to the ground or a stone dropped from the same height at the same time that the bullet is fired? I say they will both hit the grouns at the same time. Please reply.

richard - **USA**

25732

### Answer

Considering air resistance as negligible, both objects will hit the ground at the same time.

See Effect of Gravity on Sideways Motion for details.

## Calculating weight of Earth twin

Topic: **Gravitation**

### Question

November 20, 2014

I tried to calculate the weight of the earth in pounds using your method of calculation given at URL school-for-champions.com/science/gravitation_force_objects.htm but I appear to be getting the wrong answer. Could you tell me why?

I assumed there was a second earth. To weigh it, I assumed the thickness of my weight scale was zero meters so that the earth I was weighing was located touching my earth to make the distance between their mass centers equal to 2 x earth radius = 12742 km = 1.2742 X 10^7 meters.

Using your formula F = GMm/R^2 and letting M = m = 5.972 X 10^24 kg I compute a force F = 1.466 X 10^25 Newtons. This is the gravitational force applied to my weight scale. Computing the pound-force = 0.22481 X F Newtons, I have the weight of the earth equal 3.296 X 10^24 pounds.

This does NOT equal the weight of the earth converted from its known mass of 5.972 X 10^24 kg in the metric (or SI) units system. What am I doing wrong or what are you doing wrong at your referenced URL?

Sincerely yours,

Richard - **USA**

25571

### Answer

Using a twin Earth to measure its weight is an interesting exercise.

Since you already have its mass (m = 5.972 X 10^24 kg) you can simply multiply it by 9.8 m/s^2 to convert the mass to newtons and then multiply by 0.225 to get the weight in pounds.

The problem with using the equation F = GMm/R^2 to determine the force (and thus the weight) between the Earth and its twin has to do with the constraints on the equation, such that it is an approximation.

F = GMm/R^2 defines the force between two point masses at some distance. If the objects have some size, the distance between the centers must be very large compared to their size. In the case of the Earth and its twin in contact, the equation really doesn't hold, because you have to integrate the attraction over both surfaces. The math can be complex.

I went through the math and got the same answers as you did. However, it is surprising that the difference in weights is so great for this case.

See Universal Gravitation Equation for an explanation and illustration.

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