Gravitational Interactions » Truck and Car

  • Thread starter Sugi San
  • Start date
  • #1
3
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Homework Statement



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?

Homework Equations



F = Gm_1m_2/r^2



The Attempt at a Solution

 

Answers and Replies

  • #2
Doc Al
Mentor
45,032
1,331
Where are you stuck? Rearrange that formula to solve for the distance r.
 
  • #3
3
0
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
Doc Al
Mentor
45,032
1,331
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
OK.
= 1876895.86 x 10^-6
= 1.87689586 x 10^-12
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)
so how we find the distance?
Once you correctly find r^2, just take the square root.
 

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