R・R Studioの、p値=5.8e-11って何ですか?

##解決したいこと
RStudioを利用して、2群の平均の差の検定をt.test()で調べたところ、p値が5.8e-11と表示されました。
p値=5.8e-11は、いくらなのか知りたいです。
##プログラム

R
1data <- read.csv("worldseries.csv")
2natlhome <- data$natlhome
3natlaway <- data$natlaway
4
5shapiro.test(natlhome)#p値=0.1842>0.05より、正規性が成り立つ
6shapiro.test(natlaway)#p値=0.5281>0.05より、正規性が成り立つ
7
8var.test(natlhome,natlaway,alternative = "two.sided",conf.level = 0.95)
9#p値=0.1709>0.05より、等分散性が認められる
10
11t.test(natlhome,natlaway,var.equal = T,alternative = "two.sided",conf.level = 0.95)

##実行結果

> data <- read.csv("worldseries.csv")

> natlhome <- data$natlhome

> natlaway <- data$natlaway

> shapiro.test(natlhome)#p値=0.1842>0.05

	Shapiro-Wilk normality test

data:  natlhome
W = 0.97572, p-value = 0.1842


> shapiro.test(natlaway)#p値=0.5281>0.05

	Shapiro-Wilk normality test

data:  natlaway
W = 0.98445, p-value = 0.5281


> var.test(natlhome,natlaway,alternative = "two.sided",conf.level = 0.95)

	F test to compare two variances

data:  natlhome and natlaway
F = 1.3899, num df = 70, denom df = 70, p-value = 0.1709
alternative hypothesis: true ratio of variances is not equal to 1
95 percent confidence interval:
 0.8666161 2.2290084
sample estimates:
ratio of variances 
          1.389854 


> #p値=0.1709>0.05より、等分散性が認められる
> 
> t.test(natlhome,natlaway,var.equal = T,alternative = "two.sided",conf.level = 0.95)

	Two Sample t-test

data:  natlhome and natlaway
t = 7.0982, df = 140, p-value = 5.8e-11
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
 0.04540185 0.08045731
sample estimates:
mean of x mean of y 
0.6428451 0.5799155