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$We\quad know\quad that\quad F=G\quad \frac { { m }_{ 1 }\times { m }_{ 2 } }{ { x }^{ 2 } } \\ A\quad small\quad work\quad done\quad dW=Fdx\\ \int _{ R }^{ \infty }{ dW } =\int _{ R }^{ \infty }{ Fdx } \\ W=\int _{ R }^{ \infty }{ G\quad \frac { { m }_{ 1 }\times { m }_{ 2 } }{ { x }^{ 2 } } dx } =G{ m }_{ 1 }{ m }_{ 2 }\int _{ R }^{ \infty }{ \frac { 1 }{ { x }^{ 2 } } dx } \\ W=G{ m }_{ 1 }{ m }_{ 2 }{ \begin{bmatrix} \frac { -1 }{ x } \end{bmatrix} }_{ R }^{ \infty }=G{ m }_{ 1 }{ m }_{ 2 }\begin{bmatrix} \frac { -1 }{ \infty } -\frac { -1 }{ R } \end{bmatrix}\\ W=G\quad \frac { { m }_{ 1 }\times { m }_{ 2 } }{ { R } } \\ To\quad overcome\quad this\quad work\quad done\quad by\quad gravity,\quad an\quad equal\quad kinetic\quad energy\quad should\quad be\quad applied\\ K.E.=W\\ \cfrac { 1 }{ 2 } { m }_{ 2 }{ v }^{ 2 }=G\quad \frac { { m }_{ 1 }\times { m }_{ 2 } }{ { R } } ,where\quad v\quad is\quad the\quad velocity\quad of\quad the\quad body\\ { v }^{ 2 }=\frac { 2Gm }{ R } \\ v={ v }_{ e }\quad and\quad m=M\\ \therefore \boxed { { v }_{ e }=\sqrt { \frac { 2GM }{ R } } } \\ \\ To\quad find\quad the\quad escape\quad velocity\quad of\quad the\quad Sun\\ { v }_{ e }=\sqrt { \frac { 2\times 6.67408\times { 10 }^{ -11 }\times 2\times { 10 }^{ 30 } }{ 696.3\times { 10 }^{ 6 } } } \cong \boxed { 617.5\quad km/s }$