What is the gravitational force between two identical 5000 kg asteroids whose centers are apart by 100 m?

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What is the gravitational force between two identical 5000 kg asteroids whose centers are apart by 100 m?

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Answer 1 · 2016-04-13T07:10:17

$F = 1.6675 \cdot 10^{- 7} N$ $F=1.6675*10^-7N$

Explanation:

For this,you must know how to behave with Scientific Notations

Use the Newton's law of Gravitation formula

$F = G \frac{m_{1} m_{2}}{r^{2}}$ $color(brown)(F=G(m_1m_2)/(r^2)$

Here are the definition of the variables in the formula

$F = Force of Gravity$ $color(blue)("F"="Force of Gravity"$

$G = Gravitational constant = 6.67 \cdot 10^{- 11} N \frac{m^{2}}{k g^{2}}$ $color(blue)("G"="Gravitational constant"=6.67*10^-11 N(m^2)/(kg^2)$

$m_{1} = Mass of the first object = 5000 kg$ $color(blue)(m_1="Mass of the first object"="5000 kg"$

$m_{2} = Mass of second object= 5000 kg$ $color(blue)(m_2="Mass of second object=""5000 kg"$

$r = Distance between the centres of the objects = 100 m$ $color(blue)(r="Distance between the centres of the objects"="100 m"$

We need to find $F$ $F$

Before entering the values, we must convert the mass and the distance values into scientific notation (in the form of exponents)

$m_{1} = 5000 kg = 5 \cdot 1000 = 5 \cdot 10^{3}$ $color(violet)(m_1="5000 kg"=5*1000=5*10^3$ $kg$ $color(violet)("kg"$

$m_{2} = 5000 kg = 5 \cdot 1000 = 5 \cdot 10^{3}$ $color(violet)(m_2="5000 kg"=5*1000=5*10^3$ $kg$ $color(violet)("kg"$

$r = 100 m = 10^{2}$ $color(violet)(r="100 m"=10^2$ $m$ $color(violet)("m"$

Now, substitute the values in the formula

$\to F = 6.67 \cdot 10^{- 11} N \frac{m^{2}}{k g^{2}} \frac{(5 \cdot 10^{3} k g) (5 \cdot 10^{3} k g)}{{(10^{2} m)}^{2}}$ $rarrF=6.67*10^-11 N(m^2)/(kg^2)((5*10^3kg)(5*10^3kg))/((10^2m)^2)$

We have $(5 \cdot 10^{3} k g) (5 \cdot 10^{3} k g) = {(5 \cdot 10^{3} k g)}^{2}$ $(5*10^3kg)(5*10^3kg)=(5*10^3kg)^2$

$\to F = 6.67 \cdot 10^{- 11} N \frac{m^{2}}{k g^{2}} \frac{{(5 \cdot 10^{3} k g)}^{2}}{10^{4} m^{2}}$ $rarrF=6.67*10^-11 N(m^2)/(kg^2)((5*10^3kg)^2)/(10^4m^2)$

$\to F = 6.67 \cdot 10^{- 11} N \frac{m^{2}}{k g^{2}} \frac{5^{2} \cdot 10^{6} k g^{2}}{10^{4} m^{2}}$ $rarrF=6.67*10^-11 N(m^2)/(kg^2)(5^2*10^6kg^2)/(10^4m^2)$

$\to F = 6.67 \cdot 10^{- 11} N \frac{m^{2}}{k g^{2}} \frac{25 \cdot 10^{6} k g^{2}}{10^{4} m^{2}}$ $rarrF=6.67*10^-11 N(m^2)/(kg^2)(25*10^6kg^2)/(10^4m^2)$

Now we start to cancel

$\to F = 6.67 \cdot 10^{- 11} N \frac{m^{2}}{k g^{2}} \frac{25 \cdot 10^{6} k g^{2}}{10^{4} m^{2}}$ $rarrF=6.67*10^-11 N(cancel(m^2))/(cancel(kg^2))(25*10^6cancel(kg^2))/(10^4cancel(m^2))$

$\to F = 6.67 \cdot 10^{- 11} N \frac{25 \cdot 10^{6}}{10^{4}}$ $rarrF=6.67*10^-11N(25*10^6)/10^4$

$\to F = \frac{6.67 \cdot 10^{- 11} \cdot 25 \cdot 10^{6}}{10^{4}} N$ $rarrF=(6.67*10^-11*25*10^6)/10^4N$

$\to F = \frac{166.75 \cdot 10^{- 5}}{10^{4}} N$ $rarrF=(166.75*10^-5)/10^4N$

Move $10^{4}$ $10^4$ to the Numerator

Remember that,when positive exponents in the Denominator go to the Numerator,the become as Negative exponents

$\to F = 166.75 \cdot 10^{- 5} \cdot 10 - 4 N$ $rarrF=166.75*10^-5*10-4N$

$\to F = 166.75 \cdot 10^{- 9}$ $rarrF=166.75*10^-9$

$\Rightarrow F = 1.6675 \cdot 10^{- 7}$ $color(green)(rArrF=1.6675*10^-7$

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What is the gravitational force between two identical 5000 kg asteroids whose centers are apart by 100 m?

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