Is it possible to open the Settings app programatically in watchOS?

I'm currently using Swift to write an independent Apple Watch app for watchOS 7+. My app uses Location Services, and in case the user doesn't allow that at first, I'd like to have a button inside my watchOS app to send my user to the part of the Settings app (inside watchOS) where he can give the appropriate permissions for the app.

I know that in iOS/macOS you can use the openSettingsURLString string from UIApplication to do what I want, but it's not available in watchOS and I haven't found any equivalent resource inside the WKExtension class (normally the equivalent for UIApplication in the watchOS environment).

Does anyone know if there's any way to programatically open the Apple Watch Settings app from my watchOS app?

I'm asking since as far as I understand asking for Location permissions again would not work since the user has already refused to allow the permissions before, making the authorization status "denied".

I understand as well that there's the possibility of showing a modal and directing the user to go to Settings and allow the use of location services (as in "Please go to Settings > Privacy > Location Services > [AppName] to allow the use of location services") but I'd like if I could direct the user there instead as one might do in iOS.

The code below is what I'm trying to do - but (of course) it crashes since I'm not actually passing any system URL.

.alert(isPresented: $alertVisible) {
                Alert (title: Text("Please allow our app to use location services in order to get Weather Data"),
                       message: Text("Go to Settings?"),
                       primaryButton: .default(Text("Settings"), action: {
                    WKExtension.shared().openSystemURL(URL(string: "")!)
                    //UIApplication.shared.open(URL(string: UIApplication.openSettingsURLString)!)
                }),