Objects with masses of 215 kg and a 515 kg are separated by 0.300 m.?

Question

(a) Find the net gravitational force exerted by these objects on a 62.0 kg object placed midway between them.
(b)At what position (other than infinitely remote ones) can the 62.0 kg object be placed so as to experience a net force of zero? _____m from the 515 kg mass

Answer 1

The gravitotional force F of two masses m1 and m2 separated by a distance r is given by the equation

F = G * m1 * m2 / r^2

where G is a physical constant with value 6.67*10^ -11 N*m^2/kg^2

SInce the third object is placed midway then its distance from either m1 or m2 is 0.15 m.

Substituting we get
F1=3.9*10^-5 N
and
F2=9,5*10^-5 N

Since the two forces have opposite directions then
Ftotal = F1 - F2 = -5.6*10^-5 N having the direction of F2.

In order for total to be zero we have
F1=F2 ->
G*m1*m/x^2=G*m2*m/y^2 ->
m1/x^2 = m2/y^2 ->
m1*y^2=m2*x^2 (1)

where x and y the diastances of m from m1 and m2 respectively.

since

x+y=r=0.300 m

it follows that y = 0.3-x (2)
Combining (1) and (2) we have

m1*(0.3-x)^2=m2*x^2 ->
215*(0.09-0.6*x+x^2)=515x^2 ->
300x^2+129x-19,35=0

solving this we get

x1=0.12m
and
x2=-0.55 m which is of course rejected since it represents a point where there can be no equilibrium.

Answer 2

(b) You cannot achieve this position. For the net force to become zero, one or both of the masses must have a negligible or zero mass.

Answer 3

F = (m1 * m2 * 6.67E-11)/D^2 m1 is the mass of object one m2 is the mass of object two D is the distance between the centers of the two objects. F is the force of gravity between the two objects. o1 is the 3rd object. 6.67E-11 is newtons gravitational constant. (m1 * o1* 6.67E-11)/D^2 = (m2 * o1 *6.67E-11)/(.310 - D)^2 Substitute in the masses for the objects, and solve for D (almost everything will cancel).

Answer 4

Is it a problem in ur textbook..?

Answer 5

WHAT ?