Find the mass of Planet X, given that its radius is 3.39e6m and that the acceleration due to gravity on its surface is 3.73m/s2.
+26393 Find the mass of Planet X, given that its radius is 3.39e6m and that the acceleration due to gravity on its surface is 3.73m/s2.
\(\begin{array}{rcll} \text{ The mass of Planet X } &=& M \\ r &=& 3.39\cdot 10^6\ m \\ a &=& 3.73\cdot \frac{m}{s^2} \\\\ \text{Newton }\\ F &=& G \cdot \frac{ M\cdot m }{r^2} \qquad & G = 6.67\cdot 10^{-11} \cdot \frac{m^3}{kg\cdot s^2}\\ F &=& m \cdot a \\\\ F = G \cdot \frac{ M\cdot m }{r^2} &=& m \cdot a \\ G \cdot \frac{ M\cdot m }{r^2} &=& m \cdot a \\ G \cdot \frac{ M }{r^2} &=& a \\ \mathbf{M} &\mathbf{=}& \mathbf{\frac {a\cdot r^2}{G} } \\\\ M & =& \frac { 3.73\cdot \frac{m}{s^2}\cdot (3.39\cdot 10^6\ m)^2}{6.67\cdot 10^{-11} \cdot \frac{m^3}{kg\cdot s^2}} \\ M & =& \frac { 3.73\cdot \frac{m}{s^2}\cdot 3.39^2\cdot 10^{12}\ m^2}{6.67\cdot 10^{-11} \cdot \frac{m^3}{kg\cdot s^2}} \\ M & =& \frac { 3.73\cdot 3.39^2\cdot 10^{12} }{6.67\cdot 10^{-11} } \cdot \frac{m\cdot m^2\cdot kg\cdot s^2}{m^3\cdot s^2} \\ M & =& \frac { 3.73\cdot 3.39^2\cdot 10^{12}\cdot 10^{11} }{6.67} \cdot \ kg \\ M & =& \frac { 3.73\cdot 3.39^2\cdot 10^{23} }{6.67} \cdot \ kg \\ \mathbf{M} &\mathbf{=}& \mathbf{6.42661664168\cdot 10^{23} \cdot \ kg }\\ \end{array}\)
3.73=[6.67E-11 * M] / (3.39E6)^2, solve for M
Solve for M:
3.73 = 5.80399×10^-24 M
3.73 = 5.80399×10^-24 M is equivalent to 5.80399×10^-24 M = 3.73:
5.80399×10^-24 M = 3.73
Multiply both sides of 5.80399×10^-24 M = 3.73 by 1.72295×10^23:
1.72295×10^23×5.80399×10^-24 M = 1.72295×10^23×3.73
1.72295×10^23×5.80399×10^-24 = 1.:
1. M = 1.72295×10^23×3.73
1.72295×10^23×3.73 = 6.42662×10^23:
1. M = 6.42662×10^23
Multiply both sides of 1. M = 6.42662×10^23 by 1.:
1.×1. M = 1.×6.42662×10^23
1.×1. = 1.^2:
1.^2 M = 1.×6.42662×10^23
1.^2 = 1:
1 M = 1.×6.42662×10^23
1.×6.42662×10^23 = 6.42662×10^23:
Answer: |M = 6.42662×10^23 Kg.
+26393 Find the mass of Planet X, given that its radius is 3.39e6m and that the acceleration due to gravity on its surface is 3.73m/s2.
\(\begin{array}{rcll} \text{ The mass of Planet X } &=& M \\ r &=& 3.39\cdot 10^6\ m \\ a &=& 3.73\cdot \frac{m}{s^2} \\\\ \text{Newton }\\ F &=& G \cdot \frac{ M\cdot m }{r^2} \qquad & G = 6.67\cdot 10^{-11} \cdot \frac{m^3}{kg\cdot s^2}\\ F &=& m \cdot a \\\\ F = G \cdot \frac{ M\cdot m }{r^2} &=& m \cdot a \\ G \cdot \frac{ M\cdot m }{r^2} &=& m \cdot a \\ G \cdot \frac{ M }{r^2} &=& a \\ \mathbf{M} &\mathbf{=}& \mathbf{\frac {a\cdot r^2}{G} } \\\\ M & =& \frac { 3.73\cdot \frac{m}{s^2}\cdot (3.39\cdot 10^6\ m)^2}{6.67\cdot 10^{-11} \cdot \frac{m^3}{kg\cdot s^2}} \\ M & =& \frac { 3.73\cdot \frac{m}{s^2}\cdot 3.39^2\cdot 10^{12}\ m^2}{6.67\cdot 10^{-11} \cdot \frac{m^3}{kg\cdot s^2}} \\ M & =& \frac { 3.73\cdot 3.39^2\cdot 10^{12} }{6.67\cdot 10^{-11} } \cdot \frac{m\cdot m^2\cdot kg\cdot s^2}{m^3\cdot s^2} \\ M & =& \frac { 3.73\cdot 3.39^2\cdot 10^{12}\cdot 10^{11} }{6.67} \cdot \ kg \\ M & =& \frac { 3.73\cdot 3.39^2\cdot 10^{23} }{6.67} \cdot \ kg \\ \mathbf{M} &\mathbf{=}& \mathbf{6.42661664168\cdot 10^{23} \cdot \ kg }\\ \end{array}\)
