Question

given: $r = \frac{2gm}{v^2}$
$g = 6.7 \times 10^{-11}$
$m = 2.1 \times 10^{29}$
$v = 15,960,000$
find $r$.

Explanation:

Step1: Calculate \( V^2 \)

First, we need to find the square of \( V \). Given \( V = 15960000 \), we calculate \( V^2=(15960000)^2 \). Let's compute that: \( 15960000\times15960000 = (1.596\times 10^{7})^2=1.596^{2}\times10^{14}\approx2.547216\times 10^{14} \).

Step2: Calculate \( 2GM \)

Next, we calculate \( 2GM \). Given \( G = 6.7\times 10^{-11} \) and \( M = 2.1\times 10^{29} \), we substitute these values into the formula: \( 2\times G\times M=2\times6.7\times 10^{-11}\times2.1\times 10^{29} \). Using the rules of exponents (\( a^m\times a^n=a^{m + n} \)) and multiplying the coefficients: \( 2\times6.7\times2.1 = 28.14 \), and \( 10^{-11}\times10^{29}=10^{18} \), so \( 2GM = 28.14\times 10^{18}=2.814\times 10^{19} \).

Step3: Calculate \( R \)

Now, we use the formula \( R=\frac{2GM}{V^2} \). Substitute the values we found in Step1 and Step2: \( R=\frac{2.814\times 10^{19}}{2.547216\times 10^{14}} \). Using the rule of exponents for division (\( \frac{a^m}{a^n}=a^{m - n} \)): \( \frac{2.814}{2.547216}\times10^{19 - 14}\approx1.1047\times 10^{5} \).

Answer:

\( R\approx1.1047\times 10^{5} \) (or approximately 110470)