| Election Name: | 1817 Georgia gubernatorial election |
| Country: | Georgia (U.S. state) |
| Type: | Presidential |
| Ongoing: | no |
| Previous Election: | 1815 Georgia gubernatorial election |
| Previous Year: | 1815 |
| Next Election: | 1819 Georgia gubernatorial election |
| Next Year: | 1819 |
| Election Date: | 10 November 1817 |
| Nominee1: | William Rabun |
| Party1: | Democratic-Republican Party |
| Popular Vote1: | 62 |
| Percentage1: | 52.10% |
| Nominee2: | John Clark |
| Party2: | Democratic-Republican Party |
| Popular Vote2: | 57 |
| Percentage2: | 47.90% |
| Governor | |
| Before Election: | William Rabun (Acting) |
| Before Party: | Democratic-Republican Party |
| After Election: | William Rabun |
| After Party: | Democratic-Republican Party |
The 1817 Georgia gubernatorial election was held on 10 November 1817 in order to elect the Governor of Georgia. Democratic-Republican candidate and incumbent acting Governor William Rabun defeated fellow Democratic-Republican candidate John Clark in a Georgia General Assembly vote.[1]
On election day, 10 November 1817, Democratic-Republican candidate William Rabun won the election against his opponent fellow Democratic-Republican candidate John Clark. Rabun was officially sworn in as the 29th Governor of Georgia on 10 November 1817.[2]