Solve sqrt{6.67*10^-11*1.99*10^30/4.59*10^10}

$$\sqrt { \frac { 6.67 \times 10 ^ { - 11 } \times 1.99 \times 10 ^ { 30 } } { 4.59 \times 10 ^ { 10 } } }$$

Evaluate

$$\frac{1000\sqrt{67693830}}{153}\approx 53775.333493849$$

Solution Steps

To multiply powers of the same base, add their exponents. Add -11 and 30 to get 19.

$$\sqrt{\frac{6.67\times 10^{19}\times 1.99}{4.59\times 10^{10}}}$$

Cancel out 10^{10} in both numerator and denominator.

$$\sqrt{\frac{1.99\times 6.67\times 10^{9}}{4.59}}$$

Multiply 1.99 and 6.67 to get 13.2733.

$$\sqrt{\frac{13.2733\times 10^{9}}{4.59}}$$

Calculate 10 to the power of 9 and get 1000000000.

$$\sqrt{\frac{13.2733\times 1000000000}{4.59}}$$

Multiply 13.2733 and 1000000000 to get 13273300000.

$$\sqrt{\frac{13273300000}{4.59}}$$

Expand \frac{13273300000}{4.59} by multiplying both numerator and the denominator by 100.

$$\sqrt{\frac{1327330000000}{459}}$$

Rewrite the square root of the division \sqrt{\frac{1327330000000}{459}} as the division of square roots \frac{\sqrt{1327330000000}}{\sqrt{459}}.

$$\frac{\sqrt{1327330000000}}{\sqrt{459}}$$

Factor 1327330000000=1000^{2}\times 1327330. Rewrite the square root of the product \sqrt{1000^{2}\times 1327330} as the product of square roots \sqrt{1000^{2}}\sqrt{1327330}. Take the square root of 1000^{2}.

$$\frac{1000\sqrt{1327330}}{\sqrt{459}}$$

Factor 459=3^{2}\times 51. Rewrite the square root of the product \sqrt{3^{2}\times 51} as the product of square roots \sqrt{3^{2}}\sqrt{51}. Take the square root of 3^{2}.

$$\frac{1000\sqrt{1327330}}{3\sqrt{51}}$$

Rationalize the denominator of \frac{1000\sqrt{1327330}}{3\sqrt{51}} by multiplying numerator and denominator by \sqrt{51}.

$$\frac{1000\sqrt{1327330}\sqrt{51}}{3\left(\sqrt{51}\right)^{2}}$$

The square of \sqrt{51} is 51.

$$\frac{1000\sqrt{1327330}\sqrt{51}}{3\times 51}$$

To multiply \sqrt{1327330} and \sqrt{51}, multiply the numbers under the square root.

$$\frac{1000\sqrt{67693830}}{3\times 51}$$

Multiply 3 and 51 to get 153.

$$\frac{1000\sqrt{67693830}}{153}$$