Gravity aslo varies with the mass. Since this 1/87 scale planet would be 1/87 cubed of the mass of the earth, and the distance from its centre to its surface would be 1/87 of the Earth’s radius.
If earth’s mass is M, then the model planet’s mass is M/(87x87x87) = m
If earth’s radius is R, the the model planet’s radius is R/87 = r
Weight is proportional to Mass/Radius squared. Therefore W = Constant x M/(RxR)
Therefore if the prototype wagon’s weight is W, then the weight of the model wagon on the model planet would be
w = Constant x m/(rxr) = constant x { M/(87x87x87)}/{RxR/(87x87)}
= Constant x M/(RxRx87) = W/87
So on a scale planet, the weight reverts tothe linear scale of 1/87.
However, all our models are on the real planet earth, so it is acedemic.
By the way, the colume of a 1/87 scale model has EVERYTHING to do with weight, since weight is density multiplied by the volume.
This is getting too mathematical for me. I’ll think I’ll go now and just run some trains
I don’t think your reasoning is valid. Gravity at a surface is equal to the gravitation constant times the mass of the body dived by the raduis squared. The gravitation constant would be the same so if we set up a proportion it cancesl out. Mass would be 1/658503 the earth’s mass times the earths radius divided by 87 squared…wh
I’m getting confused now. I agree with what you said but am going round in circles trying to work it out. Weight is the gravitational force between two masses. The mass of the earth and the wagon are both relevant (prototype or model). If you put the numbers into your equation I think it boils down to the prototype weight divided by 87 though I am subject to correction.
What we really need is an HO scale Issac Newton to sit under an HO scale apple tree, and work it all out for us [:)]
PS Lynda, since the curvature of a 1/87 scale earth is approximately 46 miles radius, I would not worry about it when building your benchwork [(-D]
