# Gravitational Interactions » Truck and Car

1. Nov 23, 2011

### Sugi San

1. The problem statement, all variables and given/known data

i need help with this problem (Online HW)

If the force of attraction between a 1035 kg sports car and a 10680 kg cement truck is 3.930×10-6 N, how far are they apart?

2. Relevant equations

F = Gm_1m_2/r^2

3. The attempt at a solution

2. Nov 23, 2011

### Staff: Mentor

Where are you stuck? Rearrange that formula to solve for the distance r.

3. Nov 23, 2011

### Sugi San

r^2 = Gm_1m_2/F

We know m_1 = 1035, m_2 = 10680, F = 3.930 x 10^-6 and G is 6.67300 × 10^-11 m^3 kg^-1 s^-2.

So

r^2 = (6.67300 × 10^-11)(1035)(10680) / (3.930 x 10^-6)

= [(6.67300 )(1035)(10680) / (3.930)] x [10^-11/10^-6]
= [(6.67300 )(1035)(10680) / (3.930)] x 10^-5
= 18768958.6 x 10^-5
= 1876895.86 x 10^-6
= 1.87689586 x 10^-12

so how we find the distance?

4. Nov 23, 2011

### Staff: Mentor

OK.
You are messing up the exponents. When you move the decimal to the left one place you must add +1 to the exponent. You subtracted instead. (Example: 456.7 = 4.567 x 10^2)
Once you correctly find r^2, just take the square root.