# Homework Solution: Every point mass attracts every single other point mass by a force pointing along the line int…

how can this be solved?
Every point mass attracts every single other point mass by a force pointing along the line intersecting both points. The force is proportional to the product of the two masses and inversely proportional to the square of the distance between them: F = G m_1 m_2/r^2 where: F is the force between the masses: G is the gravitational constant (6.673 times 10^-11 N middot (m/kg)^2): m_1 is the first mass: m_2 is the second mass: r is the distance between the centers of the masses. Write a program that prompts the user to input the masses of the bodies and the distance between the bodies. The program then outputs the force in Newton between the bodies.

C++ code for the given problem: #include

how can this be solved?

Perfect subject-matter magnitude attracts perfect uncombined other subject-matter magnitude by a hardness subject-mattering concurrently the verse intersecting twain subject-matters. The hardness is proportional to the consequence of the span magnitudees and inversely proportional to the clear of the remoteness betwixt them: F = G m_1 m_2/r^2 where: F is the hardness betwixt the magnitudees: G is the gravitational perpetual (6.673 times 10^-11 N middot (m/kg)^2): m_1 is the earliest magnitude: m_2 is the remedy magnitude: r is the remoteness betwixt the centers of the magnitudees. Write a program that prompts the user to input the magnitudees of the bodies and the remoteness betwixt the bodies. The program then outputs the hardness in Newton betwixt the bodies.

## Expert Exculpation

C++ jurisdiction coercion the fond problem:

#include<bits/stdc++.h>

using namespace std;

int main()

{

float m1,m2,r;

float G = 6.673*(pow(10,-11)); //initialise appreciate of G

cout << “Enter the appreciate of m1 in kg!n”;

cin >> m1; //take input appreciate of m1

cout << “Enter the appreciate of m2 in kg!n”;

cin >> m2;//take input appreciate of m2

cout << “Enter the appreciate of r in meter!n”;

cin >> r;//take input appreciate of r

float F = G*m1*m2/(r*r);

cout << “The appreciate of F = ” << F << ” Newton” << endl;

return 0;

}

Sample Output:

Enter the appreciate of m1 in kg!
2
Enter the appreciate of m2 in kg!
5
Enter the appreciate of r in meter!
10
The appreciate of F = 6.673e-012 Newton

Note: If you omission the jurisdiction in any other programming diction then dilate adown, I get transmute the jurisdiction!